Exercise 1: Sums, quotients and compositions: the domain comes from the construction

16 answers to check

Thomas 1.2 builds new functions from old ones. For a sum, a difference or a product, $x$ must be in the domain of BOTH functions. For a quotient $\frac{f}{g}$ , in addition, $g(x) \ne 0$ . For a composition $(f \circ g)(x) = f(g(x))$ , $x$ must be in the domain of $g$ AND the output $g(x)$ must be in the domain of $f$ . In every case the domain is decided on the functions as they are GIVEN: a formula simplified afterwards can accept numbers that the construction refused.

Parts a) to c) use $f(x) = \sqrt{x - 1}$ and $g(x) = \sqrt{5 - x}$ . Part d) uses $h(x) = \frac{1}{x}$ and $k(x) = \frac{x}{x + 1}$ .

a) Find the domains of $f + g$ , $\frac{f}{g}$ and $\frac{g}{f}$ . Then simplify $(f \cdot f)(x)$ and give the domain of $f \cdot f$ .
b) Find $(f \circ g)(x)$ , its domain, and $(f \circ g)(1)$ .
c) Find $(g \circ f)(x)$ , its domain, and $(g \circ f)(10)$ . Is $f \circ g = g \circ f$ ?
d) Find $(k \circ h)(x)$ as a simple fraction, with its domain, and compute $(k \circ h)(2)$ .
e) Write $H(x) = \cos^3(2x - 1)$ and $G(x) = 2^{\sqrt{x + 4}}$ each as a composition of three simple functions, naming the innermost one. Give the domain of $G$ and compute $G(5)$ .

Type your answers, the page tells you right or wrong 0/16

a)

Domain of

f + g

,

Domain of

\frac{f}{g}

,

Domain of

\frac{g}{f}

,

Domain of

f \cdot f

,

b)

Domain of

f \circ g

,

(f \circ g)(1)

c)

Domain of

g \circ f

,

(g \circ f)(10)

Is

f \circ g = g \circ f

?

d)

Excluded value (smaller)

Excluded value (larger)

(k \circ h)(2)

e)

Innermost layer of

H

Outermost layer of

H

Domain of

G

,

G(5)

Show the solution

Answers

a) $f + g$ : $[1, 5]$ ; $\frac{f}{g}$ : $[1, 5)$ ; $\frac{g}{f}$ : $(1, 5]$ ; $(f \cdot f)(x) = x - 1$ with domain $[1, \infty)$
b) $(f \circ g)(x) = \sqrt{\sqrt{5 - x} - 1}$ , domain $(-\infty, 4]$ , $(f \circ g)(1) = 1$
c) $(g \circ f)(x) = \sqrt{5 - \sqrt{x - 1}}$ , domain $[1, 26]$ , $(g \circ f)(10) = \sqrt 2$ ; no
d) $(k \circ h)(x) = \frac{1}{1 + x}$ , domain $\mathbb{R} \setminus \{-1, 0\}$ , $(k \circ h)(2) = \frac{1}{3}$
e) $H = p \circ c \circ \ell$ with $\ell(x) = 2x - 1$ , $c = \cos$ , $p(u) = u^3$ ; $G = e_2 \circ r \circ s$ with $s(x) = x + 4$ , $r = \sqrt{\ }$ , $e_2(u) = 2^u$ ; domain of $G$ : $[-4, \infty)$ ; $G(5) = 8$

a)

$D(f) = [1, \infty)$ since $x - 1 \ge 0$ , and $D(g) = (-\infty, 5]$ since $5 - x \ge 0$ . The sum needs both: $[1, \infty) \cap (-\infty, 5] = [1, 5]$ . The quotient $\frac{f}{g}$ needs both AND $g(x) \ne 0$ : $\sqrt{5 - x} = 0$ at $x = 5$ , so $[1, 5)$ . The quotient $\frac{g}{f}$ loses $x = 1$ instead, where $f(1) = 0$ : $(1, 5]$ . The product $f \cdot f$ is the trap of the part: $(f \cdot f)(x) = \sqrt{x - 1}\,\sqrt{x - 1} = x - 1$ , a formula that accepts every real number, yet $(f \cdot f)(0)$ does not exist because $f(0) = \sqrt{-1}$ does not. The domain is $D(f) \cap D(f) = [1, \infty)$ , the domain of the construction, not the domain of the simplified formula $x - 1$ .

b)

$(f \circ g)(x) = f(\sqrt{5 - x}) = \sqrt{\sqrt{5 - x} - 1}$ . Inner function first: $x \le 5$ . Then its output must be a legal input of $f$ : $\sqrt{5 - x} - 1 \ge 0$ , that is $\sqrt{5 - x} \ge 1$ . Both sides are nonnegative, so squaring keeps the inequality: $5 - x \ge 1$ , $x \le 4$ . The domain is $(-\infty, 5] \cap (-\infty, 4] = (-\infty, 4]$ . Check at the edge: $(f \circ g)(4) = \sqrt{\sqrt 1 - 1} = 0$ exists, and at $x = 4.5$ the inner output is $\sqrt{0.5} \approx 0.71 < 1$ , refused by $f$ . Value: $g(1) = \sqrt 4 = 2$ , then $f(2) = \sqrt 1 = 1$ . Stopping at the first condition and answering $(-\infty, 5]$ is the lost mark: the outer root needs its own condition.

c)

$(g \circ f)(x) = g(\sqrt{x - 1}) = \sqrt{5 - \sqrt{x - 1}}$ . Inner: $x \ge 1$ . Outer: $5 - \sqrt{x - 1} \ge 0$ , so $\sqrt{x - 1} \le 5$ , and squaring two nonnegative sides, $x - 1 \le 25$ , $x \le 26$ . Domain $[1, 26]$ . Value: $f(10) = \sqrt 9 = 3$ and $g(3) = \sqrt{5 - 3} = \sqrt 2 \approx 1.4142$ . The two compositions are different functions with different domains, $(-\infty, 4]$ and $[1, 26]$ : already at $x = 10$ , $(f \circ g)(10)$ does not even exist since $g(10) = \sqrt{-5}$ . So $f \circ g \ne g \circ f$ ; the order of a composition is part of its definition.

d)

$(k \circ h)(x) = k\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{\frac{1}{x} + 1}$ , a COMPLEX FRACTION. The algebraic gesture: multiply the numerator and the denominator by $x$ , the denominator of the small fractions, which is legal because $x \ne 0$ has already been required: $\frac{1}{1 + x}$ . Domain, from the construction: $x \ne 0$ for the inner $h$ , and the output $\frac{1}{x}$ must be in the domain of $k$ , that is $\frac{1}{x} \ne -1$ , so $x \ne -1$ . The domain is $\mathbb{R} \setminus \{-1, 0\}$ . The simplified formula $\frac{1}{1 + x}$ gives $1$ at $x = 0$ , but $h(0)$ does not exist, so $(k \circ h)(0)$ does not either. Value: $h(2) = \frac{1}{2}$ , $k\left(\frac{1}{2}\right) = \frac{1/2}{3/2} = \frac{1}{3}$ , and $\frac{1}{1 + 2} = \frac{1}{3}$ agrees. Two common algebra slips here: adding $\frac{1}{x} + 1$ as $\frac{2}{x}$ , and cancelling the $\frac{1}{x}$ of the top with the one of the bottom as if the $+1$ were not there.

e)

Read $H$ the way a calculator computes it: take $x$ , form $2x - 1$ , take its cosine, cube the result. Notation first: $\cos^3(2x - 1)$ MEANS $[\cos(2x - 1)]^3$ , the cube of the cosine, never $\cos((2x - 1)^3)$ . So $\ell(x) = 2x - 1$ is innermost, then $c(u) = \cos u$ , then $p(u) = u^3$ , and $H = p \circ c \circ \ell$ , written in the reverse of the order in which it acts. For $G$ : $s(x) = x + 4$ , then $r(u) = \sqrt u$ , then $e_2(u) = 2^u$ , so $G = e_2 \circ r \circ s$ . The exponential accepts every real exponent, so only the root restricts: $x + 4 \ge 0$ , domain $[-4, \infty)$ . $G(5) = 2^{\sqrt 9} = 2^3 = 8$ . These decompositions are not unique ( $q(u) = \sqrt{u + 4}$ could be one layer), but the innermost function must act FIRST in any of them; they are exactly the layers the chain rule of chapter 10 will ask you to name.

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Exercise 2: Shifting and scaling: factor the inside, then keep the order

14 answers to check

Thomas 1.2 lists the moves. Outside $f$ : $y = f(x) + k$ shifts up $k$ , $y = cf(x)$ stretches vertically by $c$ (and reflects in the $x$ -axis if $c < 0$ ). Inside $f$ : $y = f(x - h)$ shifts RIGHT $h$ , $y = f(cx)$ COMPRESSES horizontally by $c$ for $c > 1$ . When the inside is $cx - d$ , it must be FACTORED before the shift is read: the shift is $\frac{d}{c}$ , not $d$ . Outside, the operations follow the order of the formula, multiply first, then add.

The figure shows the graph of $f$ , made of three segments through $A(-3, -1)$ , $B(0, 2)$ , $C(3, 2)$ and $D(6, 0)$ . Its domain is $[-3, 6]$ and its range is $[-1, 2]$ . We study $y = -2f(3x - 6) + 1$ .

a) Write $3x - 6$ as $3(x - h)$ . List, in a correct order, the transformations that take the graph of $f$ to the graph of $y = -2f(3x - 6) + 1$ , and give the images of $A$ , $B$ , $C$ and $D$ .
b) Give the domain and the range of $y = -2f(3x - 6) + 1$ .
c) A student compresses the graph of $f$ horizontally by $3$ , then shifts it $6$ units right. Where does her procedure send $B$ , and which function did she actually graph?
d) Compare with $y = -2\left(f(3x - 6) + 1\right)$ : image of $D$ and range.
e) Write the equation of the curve obtained from $y = \sqrt x$ by, in this order: a shift of $4$ units left, a horizontal compression by $2$ , a reflection in the $x$ -axis, a shift of $3$ units up. Give its domain and its value at $x = 6$ .

Type your answers, the page tells you right or wrong 0/14

a)

h

Image of

B

:

x

Image of

B

:

y

Image of

D

:

x

Image of

D

:

y

b)

Domain ,

Range ,

c)

Her image of

B

:

x

The function she graphed

d)

Image of

D

:

y

Range ,

e)

Smallest

x

of the domain

Value at

x = 6

Range ,

Show the solution

Answers

a) $3(x - 2)$ ; compress horizontally by $3$ , shift right $2$ , stretch vertically by $2$ and reflect in the $x$ -axis, shift up $1$ ; $A \mapsto (1, 3)$ , $B \mapsto (2, -3)$ , $C \mapsto (3, -3)$ , $D \mapsto (4, 1)$
b) Domain $[1, 4]$ , range $[-3, 3]$
c) $B \mapsto (6, 2)$ before the outside moves; she graphed $f(3(x - 6)) = f(3x - 18)$
d) $D \mapsto (4, -2)$ ; range $[-6, 0]$
e) $y = 3 - \sqrt{2x + 4}$ , domain $[-2, \infty)$ , value $-1$ at $x = 6$

a)

Factor the inside: $3x - 6 = 3(x - 2)$ , so $y = -2f(3(x - 2)) + 1$ . Inside, $3(\cdot)$ compresses horizontally by $3$ , THEN $x - 2$ shifts $2$ units right; outside, $-2$ stretches vertically by $2$ and reflects in the $x$ -axis, THEN $+1$ shifts up $1$ . The two groups act on different coordinates, so they can be done in either order with respect to each other, but not within a group. Point by point: a point $(a, b)$ of $f$ goes where the inside equals $a$ , $3x - 6 = a$ , so $x = \frac{a + 6}{3}$ , and its height becomes $-2b + 1$ . $A(-3, -1) \mapsto (1, 3)$ , $B(0, 2) \mapsto (2, -3)$ , $C(3, 2) \mapsto (3, -3)$ , $D(6, 0) \mapsto (4, 1)$ . Check one point in the formula: at $x = 2$ , $-2f(0) + 1 = -2(2) + 1 = -3$ . The solution figure shows both graphs.

b)

The domain moves with the $x$ -coordinates: $-3 \le 3x - 6 \le 6$ gives $3 \le 3x \le 12$ , so $1 \le x \le 4$ , the domain $[1, 4]$ , as the images of $A$ and $D$ confirm. The range moves with the outside only: $f$ takes every value of $[-1, 2]$ , so $-2f$ takes every value of $[-4, 2]$ (multiplying by a negative number SWAPS the ends: $-2 \times 2 = -4$ is now the smallest), and $-2f + 1$ every value of $[-3, 3]$ . Writing $[-2 \times (-1) + 1, -2 \times 2 + 1] = [3, -3]$ , an interval with its ends in the wrong order, is the typical slip.

c)

Her compression by $3$ sends $B(0, 2)$ to $(0, 2)$ , and the shift right $6$ then sends it to $(6, 2)$ , while the correct image of the inside is at $x = 2$ . Her two steps replace $x$ by $3x$ , then $x$ by $x - 6$ in the result: $f(3x)$ becomes $f(3(x - 6)) = f(3x - 18)$ , not $f(3x - 6)$ . Two correct orders exist: compress by $3$ then shift right $2$ (the factored form), or shift right $6$ then compress by $3$ ( $f(x - 6)$ becomes $f(3x - 6)$ , and $B$ goes $0 \mapsto 6 \mapsto 2$ ). The number $6$ is a shift only if it is applied BEFORE the compression; after it, the shift is $\frac{6}{3} = 2$ . The whole graph lands $4$ units too far right, which costs every point of the sketch.

d)

Now the parentheses force the addition first: $b \mapsto -2(b + 1) = -2b - 2$ . $D(6, 0)$ goes to $(4, -2)$ instead of $(4, 1)$ . The range: $f + 1$ runs over $[0, 3]$ , and $-2(f + 1)$ over $[-6, 0]$ . The two graphs have the same shape and the same domain, but the second sits $3$ units lower: $-2b - 2$ against $-2b + 1$ . Outside the parentheses the order of operations IS the order of the transformations, exactly as when you evaluate a number.

e)

Apply each step to the CURRENT formula by replacing $x$ (inside) or acting on $y$ (outside). Shift left $4$ : $\sqrt{x + 4}$ . Compress horizontally by $2$ : replace $x$ by $2x$ in that formula, $\sqrt{2x + 4}$ . Reflect in the $x$ -axis: $-\sqrt{2x + 4}$ . Up $3$ : $y = 3 - \sqrt{2x + 4}$ . Domain: $2x + 4 \ge 0$ , $x \ge -2$ , so $[-2, \infty)$ ; and at $x = 6$ , $3 - \sqrt{16} = -1$ . The trap is to write $\sqrt{2(x + 4)} = \sqrt{2x + 8}$ , which compresses FIRST and then shifts: it starts at $x = -4$ instead of $-2$ . Check with the starting point $(0, 0)$ of $\sqrt x$ : left $4$ gives $(-4, 0)$ , compressing by $2$ halves the $x$ -coordinate, $(-2, 0)$ , then $(-2, 0)$ and $(-2, 3)$ : the curve starts at $(-2, 3)$ , and indeed $3 - \sqrt{0} = 3$ .

Exercise 3: Radians, the unit circle and the six trigonometric functions

20 answers to check

An angle of $\theta$ radians cuts an arc of length $s = r\theta$ on a circle of radius $r$ ; so $\pi$ radians is $180^\circ$ , and the formula $s = r\theta$ is FALSE in degrees. On the unit circle, the terminal side of $\theta$ meets the circle at $P(\cos\theta, \sin\theta)$ ; then $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , $\sec\theta = \frac{1}{\cos\theta}$ , $\csc\theta = \frac{1}{\sin\theta}$ , $\cot\theta = \frac{\cos\theta}{\sin\theta}$ . An exact value is read in two steps: the REFERENCE ANGLE (the acute angle between the terminal side and the $x$ -axis) gives the size, the QUADRANT gives the sign.

The figure shows the unit circle and the terminal side of $\theta = \frac{4\pi}{3}$ , which meets the circle at $P$ ; the reference angle is marked in orange. A scientific calculator is allowed, but parts b) to d) must be answered exactly first.

a) Convert $150^\circ$ and $-225^\circ$ to radians, and $\frac{7\pi}{12}$ rad to degrees. Find the length of the arc cut by an angle of $150^\circ$ on a circle of radius $6$ cm.
b) Give the reference angle of $\theta = \frac{4\pi}{3}$ , then the exact values of its six trigonometric functions.
c) Given $\tan\theta = \frac{15}{8}$ with $\pi < \theta < \frac{3\pi}{2}$ , find $\sin\theta$ , $\cos\theta$ , $\sec\theta$ and $\cot\theta$ exactly.
d) Find exactly $\sin\frac{17\pi}{6}$ , $\cos\left(-\frac{10\pi}{3}\right)$ and $\tan\left(-\frac{13\pi}{4}\right)$ .
e) Compute $\sin 2$ and $\sin 2^\circ$ to four decimals. A student types $\cos(\pi \div 3)$ and reads $0.9998$ : what did the calculator compute, and what is the correct value?

Type your answers, the page tells you right or wrong 0/20

a)

150^\circ

(in multiples of

\pi

)

-225^\circ

(in multiples of

\pi

)

\frac{7\pi}{12}

rad degrees

Arc length cm

b)

Reference angle (in multiples of

\pi

)

\cos\frac{4\pi}{3}

\sin\frac{4\pi}{3}

\tan\frac{4\pi}{3}

\sec\frac{4\pi}{3}

c)

\sin\theta

\cos\theta

\sec\theta

\cot\theta

d)

\sin\frac{17\pi}{6}

\cos\left(-\frac{10\pi}{3}\right)

\tan\left(-\frac{13\pi}{4}\right)

e)

\sin 2

(radians)

\sin 2^\circ

The student's calculator was in

\cos\frac{\pi}{3}

Show the solution

Answers

a) $\frac{5\pi}{6}$ , $-\frac{5\pi}{4}$ , $105^\circ$ ; arc $5\pi \approx 15.71$ cm
b) Reference angle $\frac{\pi}{3}$ ; $\sin = -\frac{\sqrt 3}{2}$ , $\cos = -\frac{1}{2}$ , $\tan = \sqrt 3$ , $\csc = -\frac{2}{\sqrt 3}$ , $\sec = -2$ , $\cot = \frac{1}{\sqrt 3}$
c) $\sin\theta = -\frac{15}{17}$ , $\cos\theta = -\frac{8}{17}$ , $\sec\theta = -\frac{17}{8}$ , $\cot\theta = \frac{8}{15}$
d) $\frac{1}{2}$ , $-\frac{1}{2}$ , $-1$
e) $\sin 2 \approx 0.9093$ , $\sin 2^\circ \approx 0.0349$ ; it computed $\cos(1.0472^\circ)$ in degree mode; $\cos\frac{\pi}{3} = 0.5$

a)

Multiply degrees by $\frac{\pi}{180}$ : $150 \times \frac{\pi}{180} = \frac{5\pi}{6}$ and $-225 \times \frac{\pi}{180} = -\frac{5\pi}{4}$ . Multiply radians by $\frac{180}{\pi}$ : $\frac{7\pi}{12} \times \frac{180}{\pi} = 105^\circ$ . Arc length: $s = r\theta$ holds with $\theta$ IN RADIANS, so $s = 6 \times \frac{5\pi}{6} = 5\pi \approx 15.71$ cm. The answer $6 \times 150 = 900$ cm, an arc fifty times longer than the whole circle ( $12\pi \approx 37.7$ cm), is what the formula gives in degrees; a length larger than the circumference should stop you. Simplifying $\frac{150}{180}$ to $\frac{5}{6}$ before multiplying by $\pi$ keeps the arithmetic small.

b)

$\frac{4\pi}{3} = \pi + \frac{\pi}{3}$ : the terminal side has gone half a turn and then $\frac{\pi}{3}$ more, into the third quadrant, and the angle it makes with the negative $x$ -axis is the reference angle $\frac{\pi}{3}$ (orange on the figure). Size from the reference angle: $\cos\frac{\pi}{3} = \frac{1}{2}$ , $\sin\frac{\pi}{3} = \frac{\sqrt 3}{2}$ . Sign from the quadrant: in the third quadrant both coordinates of $P$ are negative. So $\cos\frac{4\pi}{3} = -\frac{1}{2}$ , $\sin\frac{4\pi}{3} = -\frac{\sqrt 3}{2} \approx -0.8660$ . Then $\tan = \frac{-\sqrt 3/2}{-1/2} = \sqrt 3 \approx 1.7321$ , positive because two negatives divide to a positive; $\sec = \frac{1}{-1/2} = -2$ ; $\csc = -\frac{2}{\sqrt 3} = -\frac{2\sqrt 3}{3} \approx -1.1547$ ; $\cot = \frac{1}{\sqrt 3}$ . The division by a fraction ( $\frac{1}{-1/2} = -2$ , not $-\frac{1}{2}$ ) is the algebra step where marks go.

c)

Reference triangle: $\tan = \frac{15}{8}$ gives legs $15$ and $8$ , and the hypotenuse is $\sqrt{15^2 + 8^2} = \sqrt{289} = 17$ . So the SIZES are $|\sin\theta| = \frac{15}{17}$ and $|\cos\theta| = \frac{8}{17}$ . The quadrant gives the signs: $\pi < \theta < \frac{3\pi}{2}$ is the third quadrant, where $\sin$ and $\cos$ are both negative and $\tan$ positive, as given. So $\sin\theta = -\frac{15}{17}$ , $\cos\theta = -\frac{8}{17}$ , $\sec\theta = -\frac{17}{8}$ , $\cot\theta = \frac{8}{15}$ . Check with the identity: $\left(\frac{15}{17}\right)^2 + \left(\frac{8}{17}\right)^2 = \frac{225 + 64}{289} = 1$ , and $\frac{\sin\theta}{\cos\theta} = \frac{15}{8}$ . Giving the positive values of the triangle loses the whole part: a triangle knows sizes, never signs.

d)

Remove full turns ( $2\pi$ for sine and cosine, $\pi$ for tangent) until the angle is in $[0, 2\pi)$ , then use a reference angle. $\frac{17\pi}{6} - 2\pi = \frac{5\pi}{6}$ , second quadrant, reference $\frac{\pi}{6}$ , sine positive: $\sin\frac{17\pi}{6} = \frac{1}{2}$ . Cosine is even, $\cos(-u) = \cos u$ : $\cos\left(-\frac{10\pi}{3}\right) = \cos\frac{10\pi}{3} = \cos\left(\frac{10\pi}{3} - 2\pi\right) = \cos\frac{4\pi}{3} = -\frac{1}{2}$ , the angle of part b). Tangent has period $\pi$ : $-\frac{13\pi}{4} + 4\pi = \frac{3\pi}{4}$ , second quadrant, reference $\frac{\pi}{4}$ , tangent negative: $-1$ . The arithmetic of fractions of $\pi$ is the step to write out: $2\pi = \frac{12\pi}{6}$ , $4\pi = \frac{16\pi}{4}$ .

e)

In radian mode, $\sin 2 \approx 0.9093$ : $2$ radians is about $114.6^\circ$ , an angle of the second quadrant, where the sine is large. In degree mode, $\sin 2^\circ \approx 0.0349$ , a tiny angle. Same keys, two different numbers. The student's calculator is in DEGREE mode: it computed $\pi \div 3 \approx 1.0472$ , then took the cosine of $1.0472$ DEGREES, which is $\approx 0.9998$ , almost $1$ because the angle is almost $0$ . The correct value is $\cos\frac{\pi}{3} = \frac{1}{2}$ , known exactly without any key. In calculus every angle is in radians, so the calculator lives in radian mode for the whole course, and a cosine of a known angle that comes out close to $1$ is the signal of the wrong mode.

Exercise 4: Identities and trigonometric equations on an interval: factor, never divide

12 answers to check

The identities of Thomas 1.3 used here: $\sin^2 x + \cos^2 x = 1$ ; $\sin 2x = 2\sin x\cos x$ ; $\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$ . To solve an equation on an interval: rewrite it with ONE function of ONE angle, bring everything to one side, FACTOR, and solve each factor with the unit circle. Two rules protect the solutions: never divide both sides by an expression that can be $0$ (it deletes the solutions where it vanishes), and when the angle is $2x$ , let $2x$ run over the DOUBLED interval.

All answers are exact multiples of $\pi$ ; no calculator key for an inverse function is needed anywhere.

a) Given $\sin\alpha = \frac{1}{3}$ with $0 < \alpha < \frac{\pi}{2}$ , find $\cos\alpha$ , $\sin 2\alpha$ and $\cos 2\alpha$ exactly.
b) Solve $2\cos 2x = 1$ on $[0, 2\pi]$ .
c) Solve $\sin 2x = \cos x$ on $[0, 2\pi]$ . A student divides both sides by $\cos x$ : what does she lose?
d) Solve $2\sin^2 x + 3\cos x = 3$ on $[0, 2\pi]$ .
e) Prove that $\cos^4 x - \sin^4 x = \cos 2x$ , then solve $\cos^4 x - \sin^4 x = -\frac{1}{2}$ on $[0, \pi]$ .

Type your answers, the page tells you right or wrong 0/12

a)

\cos\alpha

\sin 2\alpha

\cos 2\alpha

b)

Number of solutions

Smallest solution (in multiples of

\pi

)

Largest solution (in multiples of

\pi

)

c)

Number of solutions

Largest solution (in multiples of

\pi

)

d)

Number of solutions

Second smallest solution (in multiples of

\pi

)

e)

Number of solutions

Smaller solution (in multiples of

\pi

)

Show the solution

Answers

a) $\cos\alpha = \frac{2\sqrt 2}{3}$ , $\sin 2\alpha = \frac{4\sqrt 2}{9}$ , $\cos 2\alpha = \frac{7}{9}$
b) $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$
c) $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ ; dividing loses $\frac{\pi}{2}$ and $\frac{3\pi}{2}$
d) $x = 0, \frac{\pi}{3}, \frac{5\pi}{3}, 2\pi$
e) Difference of squares, then $\sin^2 x + \cos^2 x = 1$ ; $x = \frac{\pi}{3}, \frac{2\pi}{3}$

a)

Pythagoras: $\cos^2\alpha = 1 - \frac{1}{9} = \frac{8}{9}$ , so $\cos\alpha = \pm\frac{\sqrt 8}{3} = \pm\frac{2\sqrt 2}{3}$ , and the first quadrant chooses $+$ : $\cos\alpha = \frac{2\sqrt 2}{3} \approx 0.9428$ . Double angle: $\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \frac{1}{3} \cdot \frac{2\sqrt 2}{3} = \frac{4\sqrt 2}{9} \approx 0.6285$ . For $\cos 2\alpha$ the version with $\sin$ alone avoids the root: $\cos 2\alpha = 1 - 2\sin^2\alpha = 1 - \frac{2}{9} = \frac{7}{9}$ . Check: $\sin^2 2\alpha + \cos^2 2\alpha = \frac{32}{81} + \frac{49}{81} = 1$ . Two classic slips: $\sin 2\alpha = \frac{2}{3}$ (doubling the sine), and $\sqrt 8 = 4$ instead of $2\sqrt 2$ .

b)

Isolate: $\cos 2x = \frac{1}{2}$ . Name the angle $u = 2x$ : when $x$ runs over $[0, 2\pi]$ , $u$ runs over $[0, 4\pi]$ , TWO turns. On the unit circle $\cos u = \frac{1}{2}$ at $u = \frac{\pi}{3}$ and $u = \frac{5\pi}{3}$ , and again one turn later at $\frac{7\pi}{3}$ and $\frac{11\pi}{3}$ . Divide by $2$ : $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ . Four solutions, the four crossings of the solution figure. The student who solves $\cos u = \frac{1}{2}$ on $[0, 2\pi]$ only and halves finds two of them and loses half the marks; the doubled interval is written BEFORE the unit circle is used.

c)

Expand the double angle to have one angle everywhere: $2\sin x\cos x = \cos x$ . Bring everything to one side and FACTOR the common $\cos x$ : $2\sin x\cos x - \cos x = \cos x\,(2\sin x - 1) = 0$ . A product is zero when one factor is: $\cos x = 0$ gives $x = \frac{\pi}{2}, \frac{3\pi}{2}$ ; $\sin x = \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}$ . Four solutions. Dividing by $\cos x$ gives $2\sin x = 1$ and only $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ : the division was legal only where $\cos x \ne 0$ , and the lost solutions are exactly those where it was not. Check $x = \frac{\pi}{2}$ in the ORIGINAL equation: $\sin\pi = 0 = \cos\frac{\pi}{2}$ . Factoring the common factor is THE algebraic gesture of every trigonometric equation of the course, and later of every $f'(x) = 0$ .

d)

Two functions, $\sin$ and $\cos$ : use $\sin^2 x = 1 - \cos^2 x$ to keep only $\cos$ . $2(1 - \cos^2 x) + 3\cos x = 3$ becomes $-2\cos^2 x + 3\cos x - 1 = 0$ , that is $2\cos^2 x - 3\cos x + 1 = 0$ , a quadratic in $c = \cos x$ : $(2c - 1)(c - 1) = 0$ , so $c = \frac{1}{2}$ or $c = 1$ . $\cos x = \frac{1}{2}$ : $x = \frac{\pi}{3}, \frac{5\pi}{3}$ . $\cos x = 1$ : $x = 0$ AND $x = 2\pi$ , both in the CLOSED interval $[0, 2\pi]$ . Four solutions. Check $x = \frac{\pi}{3}$ : $2 \cdot \frac{3}{4} + \frac{3}{2} = 3$ . The endpoint $2\pi$ is forgotten on most copies; the brackets of the interval are part of the question.

e)

$\cos^4 x - \sin^4 x$ is a difference of SQUARES: $(\cos^2 x)^2 - (\sin^2 x)^2 = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x \cdot 1 = \cos 2x$ . The equation becomes $\cos 2x = -\frac{1}{2}$ with $u = 2x \in [0, 2\pi]$ : $u = \frac{2\pi}{3}$ or $\frac{4\pi}{3}$ , so $x = \frac{\pi}{3}$ or $\frac{2\pi}{3}$ . Check: $\cos^4\frac{\pi}{3} - \sin^4\frac{\pi}{3} = \frac{1}{16} - \frac{9}{16} = -\frac{1}{2}$ . Expanding $\cos^4 x$ as $(\cos x)^4$ and trying to solve a fourth degree equation is possible but long; recognizing the difference of squares is the rewrite that makes the question a two-line problem.

Exercise 5: The general sine function: amplitude, period and shift, from a formula and from a graph

16 answers to check

The function $y = A\sin(B(x - C)) + D$ has amplitude $|A|$ , period $\frac{2\pi}{|B|}$ , horizontal shift $C$ and midline $y = D$ ; its range is $[D - |A|, D + |A|]$ . Thomas writes the same function with the period itself in the formula, $A\sin\left(\frac{2\pi}{P}(x - C)\right) + D$ . The horizontal shift is read only after the coefficient of $x$ has been FACTORED out of the angle: $\sin(2x - \pi)$ is shifted by $\frac{\pi}{2}$ , not by $\pi$ . The same graph has many formulas: with $\cos$ instead of $\sin$ , with $A < 0$ , with $C$ changed by a period.

The figure shows the graph of a sinusoidal function $g$ , with a maximum at $(1, 5)$ and a minimum at $(4, -1)$ .

a) For $y = 3\sin\left(2x - \frac{\pi}{2}\right) + 1$ , factor the angle and give the amplitude, the period, the horizontal shift and the range.
b) From the figure, find $A > 0$ , $D$ , the period and $B > 0$ for $g(x) = A\sin(B(x - C)) + D$ , and the value of $C$ in $(-3, 3]$ .
c) Write $g$ with a cosine, $g(x) = 3\cos(B(x - C)) + 2$ , and then with $A = -3$ and a sine, giving $C$ in $(-3, 3]$ each time.
d) Solve $g(x) = 3.5$ on $[0, 6]$ .
e) The graph of $y = \sin x$ is compressed horizontally by $3$ , then shifted $\frac{\pi}{6}$ to the right. Write the new equation in the form $y = \sin(3x - c)$ . A student writes $y = \sin\left(3x - \frac{\pi}{6}\right)$ : compare the two values at $x = \frac{\pi}{3}$ .

Type your answers, the page tells you right or wrong 0/16

a)

Amplitude

Period (in multiples of

\pi

)

Shift to the right (in multiples of

\pi

)

Range ,

b)

A

D

Period

B

(in multiples of

\pi

)

C

in

(-3, 3]

c)

Cosine form:

C

Sine with

A = -3

:

C

d)

Number of solutions

The solution strictly between

0

and

6

e)

c

(in multiples of

\pi

)

Correct value at

x = \frac{\pi}{3}

The student's value at

x = \frac{\pi}{3}

Show the solution

Answers

a) $3\sin\left(2\left(x - \frac{\pi}{4}\right)\right) + 1$ : amplitude $3$ , period $\pi$ , shift $\frac{\pi}{4}$ right, range $[-2, 4]$
b) $A = 3$ , $D = 2$ , period $6$ , $B = \frac{\pi}{3}$ , $C = -\frac{1}{2}$
c) $g(x) = 3\cos\left(\frac{\pi}{3}(x - 1)\right) + 2$ ; $g(x) = -3\sin\left(\frac{\pi}{3}\left(x - \frac{5}{2}\right)\right) + 2$
d) $x = 0$ , $x = 2$ , $x = 6$
e) $y = \sin\left(3x - \frac{\pi}{2}\right)$ , so $c = \frac{\pi}{2}$ ; at $\frac{\pi}{3}$ : $1$ against the student's $\frac{1}{2}$

a)

Factor the $2$ out of the angle: $2x - \frac{\pi}{2} = 2\left(x - \frac{\pi}{4}\right)$ . So $A = 3$ , $B = 2$ , $C = \frac{\pi}{4}$ , $D = 1$ : amplitude $3$ , period $\frac{2\pi}{2} = \pi$ , shift $\frac{\pi}{4}$ to the RIGHT, midline $y = 1$ , range $[1 - 3, 1 + 3] = [-2, 4]$ . Check: the sine starts its cycle where the angle is $0$ , $2x - \frac{\pi}{2} = 0$ , that is $x = \frac{\pi}{4}$ , and there $y = 3\sin 0 + 1 = 1$ , on the midline, going up. Reading the shift as $\frac{\pi}{2}$ from the unfactored angle is the mark lost on nearly every first attempt.

b)

Midline halfway between the extremes: $D = \frac{5 + (-1)}{2} = 2$ . Amplitude: $A = \frac{5 - (-1)}{2} = 3$ . From a maximum to the next minimum is HALF a period: $4 - 1 = 3$ , so the period is $6$ and $B = \frac{2\pi}{6} = \frac{\pi}{3}$ . For $C$ : $\sin$ starts a cycle on the midline going UP, a quarter period ( $1.5$ ) before the maximum: at $x = 1 - 1.5 = -0.5$ . So $C = -\frac{1}{2}$ and $g(x) = 3\sin\left(\frac{\pi}{3}\left(x + \frac{1}{2}\right)\right) + 2$ . Check at the maximum: $\frac{\pi}{3} \cdot \frac{3}{2} = \frac{\pi}{2}$ and $3\sin\frac{\pi}{2} + 2 = 5$ . Taking the period as $3$ , the distance between a maximum and a minimum, doubles $B$ and draws two cycles where the figure shows one.

c)

A cosine starts its cycle at a MAXIMUM, so it needs no quarter-period shift: $C = 1$ , $g(x) = 3\cos\left(\frac{\pi}{3}(x - 1)\right) + 2$ . A sine with $A = -3$ is reflected, so it starts on the midline going DOWN; after the maximum at $1$ the curve crosses the midline downward a quarter period later, at $x = 2.5$ : $g(x) = -3\sin\left(\frac{\pi}{3}\left(x - \frac{5}{2}\right)\right) + 2$ . Check at $x = 4$ : $-3\sin\frac{\pi}{2} + 2 = -1$ , the minimum. Three formulas, one graph: $C$ depends on WHICH function starts the cycle, and adding the period $6$ to any $C$ gives yet another correct formula, which is why the question fixes the interval $(-3, 3]$ .

d)

Use the cosine form: $3\cos\left(\frac{\pi}{3}(x - 1)\right) + 2 = 3.5$ gives $\cos u = \frac{1}{2}$ with $u = \frac{\pi}{3}(x - 1)$ . For $x \in [0, 6]$ , $u$ runs over $\left[-\frac{\pi}{3}, \frac{5\pi}{3}\right]$ , and there $\cos u = \frac{1}{2}$ at $u = -\frac{\pi}{3}$ , $u = \frac{\pi}{3}$ and $u = \frac{5\pi}{3}$ . Back to $x = 1 + \frac{3u}{\pi}$ : $x = 0$ , $x = 2$ , $x = 6$ . Three solutions, two of them at the ENDS of the interval. Check: $g(0) = 3\cos\left(-\frac{\pi}{3}\right) + 2 = 3.5$ and $g(6) = 3\cos\frac{5\pi}{3} + 2 = 3.5$ . Solving $\cos u = \frac{1}{2}$ on $[0, 2\pi]$ out of habit finds only $x = 2$ and $x = 6$ : the interval for $u$ must be computed from the interval for $x$ .

e)

Apply each move to the current formula. Compress by $3$ : $\sin 3x$ . Shift right $\frac{\pi}{6}$ : replace $x$ by $x - \frac{\pi}{6}$ , $\sin\left(3\left(x - \frac{\pi}{6}\right)\right) = \sin\left(3x - \frac{\pi}{2}\right)$ , so $c = \frac{\pi}{2}$ . At $x = \frac{\pi}{3}$ : $\sin\left(\pi - \frac{\pi}{2}\right) = \sin\frac{\pi}{2} = 1$ . The student's formula gives $\sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{5\pi}{6} = \frac{1}{2}$ : his graph is shifted by only $\frac{\pi}{18}$ , since $3x - \frac{\pi}{6} = 3\left(x - \frac{\pi}{18}\right)$ . The shift multiplies by the coefficient when it goes inside: $3 \times \frac{\pi}{6} = \frac{\pi}{2}$ . It is the gesture of part a), run backwards.

Exercise 6: Exponential functions and the laws of exponents: rewrite as one power

17 answers to check

For $a > 0$ and all real $x$ , $y$ : $a^x a^y = a^{x + y}$ , $\frac{a^x}{a^y} = a^{x - y}$ , $(a^x)^y = a^{xy}$ , $(ab)^x = a^x b^x$ , $a^{-x} = \frac{1}{a^x}$ , and $a^{m/n} = \sqrt[n]{a^m}$ . The exponential function $f(x) = a^x$ ( $a > 0$ , $a \ne 1$ ) is defined for every real $x$ , takes only positive values, and increases when $a > 1$ ; the natural exponential uses $e \approx 2.71828$ . The laws hold for PRODUCTS and QUOTIENTS of powers, never for sums: nothing simplifies $2^x + 2^{3x}$ except a common factor.

The figure shows the graphs of $y = 2^x$ and $y = x^2$ for $-2 \le x \le 5$ . No logarithm is needed anywhere in this exercise.

a) For $x > 0$ , write $\frac{\sqrt x\,\left(x^{-2/3}\right)^3}{\sqrt[4]{x^6}}$ as a single power of $x$ , and write $\frac{x^2 + \sqrt x}{x^{3/2}}$ as a sum of two powers of $x$ .
b) Write $\frac{(9e^{4x})^{1/2}}{(27e^{-3x})^{1/3}}$ in the form $Ce^{kx}$ , and $\frac{e^{3x} - e^x}{e^x}$ in the form $e^{kx} - 1$ .
c) Let $f(x) = 3^x$ . Show that $f(x + 2) = 9f(x)$ , factor $f(x + h) - f(x)$ , and show that $f(2x) = [f(x)]^2$ . Evaluate $f(x + h) - f(x)$ for $x = 2$ , $h = 1$ .
d) Show that the graph of $y = 2^{x + 3}$ is a vertical stretch of the graph of $y = 2^x$ , and that $y = \left(\frac{1}{4}\right)^x$ is a reflection and a horizontal compression of it; give both factors. Where do $y = 2^{x + 3}$ and $y = \left(\frac{1}{4}\right)^x$ meet?
e) From the figure and a few exact values, compare $2^x$ and $x^2$ at $x = 3$ and at $x = 10$ , find the two positive values where they are equal, and say how many times the two graphs meet.

Type your answers, the page tells you right or wrong 0/17

a)

Exponent of the single power

Sum: larger exponent

Sum: smaller exponent

b)

C

k

in

Ce^{kx}

k

in

e^{kx} - 1

c)

\frac{f(x + 2)}{f(x)}

f(3) - f(2)

f(2 \cdot 1.5)

d)

Vertical stretch factor

Horizontal compression factor

Meeting point:

x

Meeting point:

y

e)

Larger at

x = 3

Larger at

x = 10

Largest

x

where they are equal

Number of meeting points

Show the solution

Answers

a) $x^{-3}$ ; $x^{1/2} + x^{-1}$
b) $e^{3x}$ ( $C = 1$ , $k = 3$ ); $e^{2x} - 1$
c) $3^{x + 2} = 3^2 \cdot 3^x$ ; $f(x + h) - f(x) = 3^x(3^h - 1)$ ; $3^{2x} = (3^x)^2$ ; value $18$
d) $2^{x + 3} = 8 \cdot 2^x$ : stretch by $8$ ; $\left(\frac{1}{4}\right)^x = 2^{-2x}$ : reflection in the $y$ -axis and compression by $2$ ; they meet at $(-1, 4)$
e) At $3$ : $8 < 9$ ; at $10$ : $1024 > 100$ ; equal at $x = 2$ and $x = 4$ ; three meeting points (the third at a negative $x$ )

a)

Rewrite every root as a power FIRST: $\sqrt x = x^{1/2}$ , $\left(x^{-2/3}\right)^3 = x^{-2}$ (multiply the exponents), $\sqrt[4]{x^6} = x^{6/4} = x^{3/2}$ . Then add and subtract the exponents: $x^{1/2 - 2 - 3/2} = x^{-3} = \frac{1}{x^3}$ . Check at $x = 4$ : $\frac{2 \cdot 4^{-2}}{4^{3/2}} = \frac{2/16}{8} = \frac{1}{64} = 4^{-3}$ . For the second expression, SPLIT the fraction over its denominator, which is allowed because the sum is in the numerator: $\frac{x^2}{x^{3/2}} + \frac{x^{1/2}}{x^{3/2}} = x^{1/2} + x^{-1}$ . Cancelling the $x^2$ with part of the denominator while forgetting the $\sqrt x$ is the slip. This rewriting, one power at a time, is exactly what the power rule of chapter 8 will need before it can be used.

b)

A power of a product is the product of the powers: $(9e^{4x})^{1/2} = 9^{1/2}\,(e^{4x})^{1/2} = 3e^{2x}$ , and $(27e^{-3x})^{1/3} = 27^{1/3}\,(e^{-3x})^{1/3} = 3e^{-x}$ . The quotient is $\frac{3e^{2x}}{3e^{-x}} = e^{2x - (-x)} = e^{3x}$ : $C = 1$ , $k = 3$ . The minus sign of $-(-x)$ is where the answer $e^{x}$ comes from. For the second, split: $\frac{e^{3x}}{e^x} - \frac{e^x}{e^x} = e^{2x} - 1$ . Writing $\frac{e^{3x} - e^x}{e^x} = e^{3x} - 1$ , cancelling one $e^x$ from one term only, is the error a quick check at $x = 1$ catches: $\frac{e^3 - e}{e} = e^2 - 1 \approx 6.39$ , not $e^3 - 1 \approx 19.09$ .

c)

$f(x + 2) = 3^{x + 2} = 3^x \cdot 3^2 = 9 \cdot 3^x = 9f(x)$ : adding $2$ to the input MULTIPLIES the output by $9$ , the signature of an exponential. $f(x + h) - f(x) = 3^x 3^h - 3^x = 3^x(3^h - 1)$ , by factoring the common factor $3^x$ ; it is the factorization that the derivative of an exponential will start from. $f(2x) = 3^{2x} = (3^x)^2 = [f(x)]^2$ . At $x = 2$ , $h = 1$ : $3^3 - 3^2 = 27 - 9 = 18$ , and $3^2(3^1 - 1) = 9 \cdot 2 = 18$ . The trap is to treat $f$ like a linear function: $f(x + h) = f(x) + f(h)$ would give $3^3 = 3^2 + 3^1 = 12$ , false.

d)

$2^{x + 3} = 2^3 \cdot 2^x = 8 \cdot 2^x$ : the shift of $3$ to the left IS a vertical stretch by $8$ , a coincidence that only exponential functions have. $\left(\frac{1}{4}\right)^x = (2^{-2})^x = 2^{-2x}$ : inside, $x$ is replaced by $-2x$ , a reflection in the $y$ -axis and a horizontal compression by $2$ . To find where they meet, write both sides in base $2$ : $2^{x + 3} = 2^{-2x}$ . An exponential with base $2 \ne 1$ takes each value once, so the exponents are equal: $x + 3 = -2x$ , $x = -1$ , and $y = 2^{2} = 4$ . Check: $\left(\frac{1}{4}\right)^{-1} = 4$ . Bringing both sides to ONE base is the whole method; comparing $2^{x+3}$ and $4^{-x}$ without it leads nowhere.

e)

At $x = 3$ : $2^3 = 8 < 3^2 = 9$ , so the power is ahead. At $x = 10$ : $2^{10} = 1024 > 10^2 = 100$ , and far ahead: each step of $1$ doubles $2^x$ but multiplies $x^2$ by only $\left(\frac{x + 1}{x}\right)^2$ , which is less than $2$ once $x \ge 3$ . Equality for positive $x$ : $2^2 = 4 = 2^2$ and $2^4 = 16 = 4^2$ , so $x = 2$ and $x = 4$ , the two crossings on the right of the figure. For negative $x$ , $2^x$ is small and positive while $x^2$ grows: at $x = -1$ , $2^{-1} - 1 = -\frac{1}{2} < 0$ ; at $x = -\frac{1}{2}$ , $2^{-1/2} - \frac{1}{4} \approx 0.46 > 0$ ; the difference changes sign, the curves cross between $-1$ and $-\frac{1}{2}$ , as the figure shows. Three meeting points in all, and for $x > 4$ the exponential stays above for good.

Exercise 7: Period, range and domain of combined trigonometric and exponential functions

18 answers to check

A function $f$ is periodic with period $p > 0$ if $f(x + p) = f(x)$ for every $x$ of its domain; THE period is the smallest such $p$ . $\sin$ and $\cos$ have period $2\pi$ , $\tan$ has period $\pi$ , and $\sin(Bx)$ has period $\frac{2\pi}{|B|}$ . Before reading a period, a range or a domain, rewrite: a square of a sine is a cosine of the double angle, $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$ ; a composition is read from the inside, the inner function deciding which values the outer one receives.

The figure shows $y = \sin^2 x$ (solid) and $y = \sin x$ (dashed) on $[-\pi, 2\pi]$ .

a) Read the period and the range of $y = \sin^2 x$ on the figure. Prove them by writing $\sin^2 x$ in the form $A\cos(Bx) + D$ .
b) Find the domain of $h(x) = \sqrt{\cos x}$ within $[0, 2\pi]$ , then the range and the period of $q(x) = 2^{\sin x}$ .
c) Find the period of $\sin 3x$ , of $\cos 2x$ , and of $k(x) = \sin 3x + \cos 2x$ . Prove that no positive number smaller than your answer is a period of $k$ .
d) Find the periods of $|\sin x|$ , of $\cos^2 3x$ and of $\tan\frac{x}{2}$ .
e) Which of $x\sin x$ , $\sin(x^2)$ and $2^{\cos x}$ are periodic? Justify each answer in one line.

Type your answers, the page tells you right or wrong 0/18

a)

Period (in multiples of

\pi

)

Range ,

A

B

D

b)

First interval of the domain of

h

(in multiples of

\pi

) ,

Second interval (in multiples of

\pi

) ,

Range of

q

,

Period of

q

(in multiples of

\pi

)

c)

Period of

\sin 3x

(in multiples of

\pi

)

Period of

\cos 2x

(in multiples of

\pi

)

Period of

k

(in multiples of

\pi

)

d)

|\sin x|

(in multiples of

\pi

)

\cos^2 3x

(in multiples of

\pi

)

\tan\frac{x}{2}

(in multiples of

\pi

)

e)

x\sin x

periodic?

\sin(x^2)

periodic?

2^{\cos x}

periodic?

Show the solution

Answers

a) Period $\pi$ , range $[0, 1]$ ; $\sin^2 x = -\frac{1}{2}\cos 2x + \frac{1}{2}$
b) $\left[0, \frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2}, 2\pi\right]$ ; range of $q$ : $\left[\frac{1}{2}, 2\right]$ , period $2\pi$
c) $\frac{2\pi}{3}$ , $\pi$ , and $2\pi$ for $k$ : its minimum $-2$ is reached only at $\frac{\pi}{2} + 2j\pi$
d) $\pi$ , $\frac{\pi}{3}$ , $2\pi$
e) $x\sin x$ : no; $\sin(x^2)$ : no; $2^{\cos x}$ : yes, period $2\pi$

a)

On the figure the solid curve repeats between consecutive zeros $0$ , $\pi$ , $2\pi$ : period $\pi$ , half the period of $\sin x$ , because squaring turns every negative arch of the dashed curve into a positive one. Range $[0, 1]$ . Proof: $\cos 2x = 1 - 2\sin^2 x$ gives $\sin^2 x = \frac{1 - \cos 2x}{2} = -\frac{1}{2}\cos 2x + \frac{1}{2}$ : $A = -\frac{1}{2}$ , $B = 2$ , $D = \frac{1}{2}$ . So the period is $\frac{2\pi}{2} = \pi$ , the midline $y = \frac{1}{2}$ , the amplitude $\frac{1}{2}$ , and the range $\left[\frac{1}{2} - \frac{1}{2}, \frac{1}{2} + \frac{1}{2}\right] = [0, 1]$ . Reading the period $2\pi$ of $\sin x$ off the formula $\sin^2 x$ is the trap; the rewrite shows the angle $2x$ hidden in the square.

b)

$\sqrt{\cos x}$ needs $\cos x \ge 0$ : on $[0, 2\pi]$ , the cosine (the $x$ -coordinate on the unit circle) is nonnegative in the first and fourth quadrants, including their edges, so the domain is $\left[0, \frac{\pi}{2}\right] \cup \left[\frac{3\pi}{2}, 2\pi\right]$ . For $q(x) = 2^{\sin x}$ , read inside out: $\sin x$ takes every value of $[-1, 1]$ , and $2^u$ is increasing, so $q$ takes every value from $2^{-1} = \frac{1}{2}$ to $2^1 = 2$ : range $\left[\frac{1}{2}, 2\right]$ . Period: $q(x + 2\pi) = 2^{\sin(x + 2\pi)} = q(x)$ , and no smaller period works because $q = 2$ only where $\sin x = 1$ , at $\frac{\pi}{2} + 2j\pi$ , $2\pi$ apart. Giving the range $[-2, 2]$ , as if $2^{\sin x}$ were $2\sin x$ , confuses an exponent with a factor.

c)

$\sin 3x$ has period $\frac{2\pi}{3}$ and $\cos 2x$ has period $\pi$ . A period of the sum must bring BOTH back, so it must be a common multiple of $\frac{2\pi}{3}$ and $\pi$ : $m \cdot \frac{2\pi}{3} = n\pi$ needs $2m = 3n$ , first with $m = 3$ , $n = 2$ , which gives $2\pi$ . So $2\pi$ is a period of $k$ . Smaller ones are ruled out by a value reached rarely: $k(x) = -2$ needs $\sin 3x = -1$ AND $\cos 2x = -1$ . The second gives $x = \frac{\pi}{2} + n\pi$ , the first $x = \frac{\pi}{2} + \frac{2m\pi}{3}$ ; both hold when $n\pi = \frac{2m\pi}{3}$ , that is at $x = \frac{\pi}{2} + 2j\pi$ only. The minimum is reached exactly once every $2\pi$ , so no period is smaller: the period is $2\pi$ . Answering $\pi$ (the larger of the two periods) fails at once: $k\left(\frac{\pi}{2}\right) = -2$ while $k\left(\frac{3\pi}{2}\right) = \sin\frac{9\pi}{2} + \cos 3\pi = 0$ .

d)

$|\sin(x + \pi)| = |-\sin x| = |\sin x|$ , and $|\sin x| = 0$ only at multiples of $\pi$ , so the period is $\pi$ . Rewrite the square: $\cos^2 3x = \frac{1 + \cos 6x}{2}$ , period $\frac{2\pi}{6} = \frac{\pi}{3}$ , not the $\frac{2\pi}{3}$ of $\cos 3x$ . For the tangent the base period is $\pi$ , not $2\pi$ : $\tan\frac{x}{2}$ has period $\frac{\pi}{1/2} = 2\pi$ . Two formulas are mixed on many copies here: $\frac{2\pi}{|B|}$ for $\sin$ and $\cos$ , $\frac{\pi}{|B|}$ for $\tan$ .

e)

$x\sin x$ : no. Its zeros are the multiples of $\pi$ , but its values grow without bound in size, for instance $x\sin x = x$ at $x = \frac{\pi}{2} + 2j\pi$ , so no shift can bring the graph back onto itself. $\sin(x^2)$ : no. Its zeros are at $x = \sqrt{j\pi}$ , and the gaps between them, $\sqrt{(j + 1)\pi} - \sqrt{j\pi}$ , shrink as $j$ grows, while a periodic function repeats its zeros at equal spacing. $2^{\cos x}$ : yes, period $2\pi$ , since the inner function has that period and $2^u$ takes each value once; composing ANY function after a periodic inner function keeps the result periodic, while composing a periodic function after $x^2$ does not.

Exercise 8: Five statements to correct

10 answers to check

Each statement below was written by a student on a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample with numbers, and write the correct statement.

a) $\sin 2x = 2\sin x$ for every $x$ .
b) $\cos^2 x$ and $\cos(x^2)$ are two ways of writing the same function.
c) $\sin 30 = \frac{1}{2}$ .
d) The domain of $\frac{f}{g}$ is the set of numbers that are in the domain of $f$ and in the domain of $g$ .
e) $x^{-2/3} = -\sqrt[3]{x^2}$ for every $x > 0$ .

Show the solution

Answers

a) False: at $x = \frac{\pi}{2}$ , $\sin\pi = 0 \ne 2$ . $\sin 2x = 2\sin x\cos x$ .
b) False: at $x = \pi$ , $\cos^2\pi = 1$ but $\cos(\pi^2) \approx -0.9027$ . $\cos^2 x = (\cos x)^2$ .
c) False: $\sin 30 \approx -0.9880$ (radians). $\sin 30^\circ = \sin\frac{\pi}{6} = \frac{1}{2}$ .
d) False: $\frac{x}{x^2 - 9}$ is undefined at $\pm 3$ . Remove also the zeros of $g$ .
e) False: $8^{-2/3} = \frac{1}{4}$ , not $-4$ . $x^{-2/3} = \frac{1}{\sqrt[3]{x^2}}$ .

a)

FALSE. At $x = \frac{\pi}{2}$ : $\sin 2x = \sin\pi = 0$ , while $2\sin\frac{\pi}{2} = 2$ . Doubling the ANGLE does not double the sine: the sine of an angle is at most $1$ , so $2\sin x$ , which reaches $2$ , cannot be a sine at all. Correct statement: $\sin 2x = 2\sin x\cos x$ for every $x$ . The two expressions agree only where $2\sin x\cos x = 2\sin x$ , that is $\sin x\,(\cos x - 1) = 0$ , at the multiples of $\pi$ , which is why a check at $x = 0$ does not catch the error.

b)

FALSE. $\cos^2 x$ means $(\cos x)^2$ : first the cosine, then the square. $\cos(x^2)$ squares first, then takes the cosine. At $x = \pi$ : $(\cos\pi)^2 = (-1)^2 = 1$ , while $\cos(\pi^2) \approx \cos(9.8696) \approx -0.9027$ . The two functions do not even share a period: $\cos^2 x$ has period $\pi$ , and $\cos(x^2)$ is not periodic. Correct statement: $\cos^2 x = (\cos x)^2$ , a composition whose OUTER layer is the square; the same convention holds for $\sin^3 x$ , $\tan^2 x$ , but NOT for $\sin^{-1} x$ , which chapter 13 will define differently.

c)

FALSE as written. Without a degree sign, $30$ is $30$ RADIANS, almost five turns: $\sin 30 \approx -0.9880$ , which is what a calculator in radian mode returns. $30$ radians is $30 \times \frac{180}{\pi} \approx 1718.9^\circ$ , and $1718.9 - 4 \times 360 = 278.9^\circ$ , an angle of the fourth quadrant where the sine is negative, as found. Correct statement: $\sin 30^\circ = \sin\frac{\pi}{6} = \frac{1}{2}$ . In calculus a number with no unit is in radians, always.

d)

FALSE. Take $f(x) = x$ and $g(x) = x^2 - 9$ : both are defined on all of $\mathbb{R}$ , yet $\frac{f}{g}(3) = \frac{3}{0}$ does not exist, nor does $\frac{f}{g}(-3)$ . Correct statement (Thomas 1.2): the domain of $\frac{f}{g}$ is the set of $x$ in the domain of $f$ and in the domain of $g$ for which $g(x) \ne 0$ . Here, $\mathbb{R} \setminus \{-3, 3\}$ : two values removed, found by FACTORING $x^2 - 9 = (x - 3)(x + 3)$ , not only $x = 3$ .

e)

FALSE. A negative exponent means a RECIPROCAL, never a negative number: $x^{-2/3} = \frac{1}{x^{2/3}} = \frac{1}{\sqrt[3]{x^2}}$ . Counterexample $x = 8$ : $8^{2/3} = (\sqrt[3]8)^2 = 4$ , so $8^{-2/3} = \frac{1}{4}$ , while the statement gives $-4$ . And for $x > 0$ every power $x^r$ is POSITIVE, whatever the sign of $r$ , so a negative right-hand side could never be right. This is the rewrite on which the power rule depends: $\frac{1}{\sqrt[3]{x^2}}$ must become $x^{-2/3}$ before chapter 8 can differentiate it.

Exercise 9: A tide modelled by a sinusoid: when can the ship leave?

13 answers to check

In a harbour, on a given day, high tide brings the water to a depth of $7.4$ m at 4:00, and the next low tide, $1.8$ m, comes at 10:12. Between them the depth follows a sinusoid. Let $h(t)$ be the depth in metres, $t$ hours after midnight (so 10:12 is $t = 10.2$ ), and model it by $h(t) = A\cos(B(t - C)) + D$ with $A > 0$ , $B > 0$ .

The figure shows the model over one day. Calculator in RADIAN mode; the times asked in d) come out of exact values of the unit circle.

a) Find $D$ , $A$ , the period and $B$ .
b) Explain why $C = 4$ is a correct choice, write the model, and compute the depth at 8:00 to the nearest centimetre.
c) Find the times of the next high tide and of the low tide after it.
d) A ship needs at least $3.2$ m of water. Between 4:00 and the next high tide, find exactly when the depth is below $3.2$ m, and give the times to the minute.
e) If the model keeps its period, at what time is the first high tide of the next day? By how many minutes does high tide move each day?

Show the solution

Answers

a) $D = 4.6$ , $A = 2.8$ , period $12.4$ h, $B = \frac{2\pi}{12.4} \approx 0.5067$
b) The cosine starts at its maximum, at $t = 4$ ; $h(t) = 2.8\cos\left(\frac{2\pi}{12.4}(t - 4)\right) + 4.6$ ; $h(8) \approx 3.37$ m
c) High tide at $t = 16.4$ (16:24), low tide at $t = 22.6$ (22:36)
d) From $t = 4 + \frac{12.4}{3} \approx 8.13$ (8:08) to $t = 4 + \frac{24.8}{3} \approx 12.27$ (12:16), about $4.13$ h
e) At $t = 28.8$ , that is 4:48 the next day; high tide moves $48$ minutes later each day

a)

Midline halfway between high and low: $D = \frac{7.4 + 1.8}{2} = 4.6$ m. Amplitude, half the range: $A = \frac{7.4 - 1.8}{2} = 2.8$ m. From a high tide to the next low tide is HALF a period: $10.2 - 4 = 6.2$ h, so the period is $12.4$ h. Then $B = \frac{2\pi}{12.4} = \frac{\pi}{6.2} \approx 0.5067$ per hour. Converting 10:12 as $10.12$ instead of $10 + \frac{12}{60} = 10.2$ is the unit slip that shifts everything: minutes are sixtieths of an hour, not hundredths.

b)

A cosine $A\cos(B(t - C)) + D$ with $A > 0$ is at its MAXIMUM when the angle is $0$ , at $t = C$ . The first maximum is at $t = 4$ , so $C = 4$ : $h(t) = 2.8\cos\left(\frac{2\pi}{12.4}(t - 4)\right) + 4.6$ . Check the low tide: $h(10.2) = 2.8\cos\pi + 4.6 = 1.8$ . At 8:00: the angle is $\frac{2\pi}{12.4} \times 4 \approx 2.0268$ rad, $\cos 2.0268 \approx -0.4404$ , so $h(8) \approx 4.6 - 1.2331 = 3.37$ m. In degree mode the calculator returns $\cos(2.0268^\circ) \approx 0.9994$ and a depth of $7.40$ m, the high tide four hours after it has passed: the wrong mode gives a plausible but false number, which is why the mode is checked first.

c)

The maxima repeat every period: $4 + 12.4 = 16.4$ , that is 16:24 (since $0.4$ h $= 24$ min). The minima are half a period after each maximum: $16.4 + 6.2 = 22.6$ , that is 22:36. On the figure, these are the third and fourth marked points. Adding $12.4$ to 4:00 as if it were a time, 4:00 plus 12:40, gives 16:40, the base- $60$ slip again.

d)

Solve $h(t) = 3.2$ : $2.8\cos u + 4.6 = 3.2$ with $u = \frac{2\pi}{12.4}(t - 4)$ , so $\cos u = \frac{-1.4}{2.8} = -\frac{1}{2}$ . Between the high tides at $t = 4$ and $t = 16.4$ , $u$ runs over $[0, 2\pi]$ , and $\cos u = -\frac{1}{2}$ at $u = \frac{2\pi}{3}$ and $u = \frac{4\pi}{3}$ , a third and two thirds of the cycle. So $t - 4 = \frac{12.4}{3} \approx 4.133$ and $t - 4 = \frac{2 \times 12.4}{3} \approx 8.267$ : $t \approx 8.133$ (8:08) and $t \approx 12.267$ (12:16). Between them $\cos u < -\frac{1}{2}$ and the depth is below $3.2$ m: the ship cannot move from 8:08 to 12:16, about $4.13$ h, $4$ h $8$ min. The exact angles are what make the times exact; a solution that only tests a few hours cannot give them to the minute.

e)

Each high tide comes $12.4$ h after the previous one: $4$ , $16.4$ , $28.8$ . The third one is $28.8 - 24 = 4.8$ h after the next midnight, that is 4:48 the next day. Two periods, $24.8$ h, exceed a day by $0.8$ h $= 48$ min: high tide comes $48$ minutes later each day, which is why tide tables must be printed day by day. The model is idealized (real tides also vary in height through the month), but its period is the part that holds best.

Exercise 10: A final exam problem: fuel economy, a composition of unit conversions

12 answers to check

Fuel economy is quoted in miles per US gallon (mpg) in the United States, in litres per $100$ km in Canada. Use $1$ mile $= 1.609344$ km and $1$ US gallon $= 3.785412$ L. Let $k(m)$ be the distance in km travelled per litre by a car that does $m$ mpg, and $h(E) = \frac{100}{E}$ the consumption in litres per $100$ km of a car that travels $E$ km per litre.

The figure shows the consumption $c = h(k(m))$ in L per $100$ km against $m$ , with the two cars of part c). Calculator allowed.

a) Show that $k(m) = \frac{1.609344}{3.785412}\,m$ , then compute $k(30)$ and $h(k(30))$ to two decimals.
b) Write $c(m) = (h \circ k)(m)$ as $\frac{K}{m}$ , giving $K$ to two decimals, and the domain of the model. Does $k(h(30))$ mean anything?
c) Car A improves from $15$ to $20$ mpg, car B from $40$ to $50$ mpg. For each, find the fuel saved per $100$ km, and say which improvement saves more.
d) To what fuel economy would car B have to go from $40$ mpg to save as much fuel per $100$ km as car A? Give an exact answer.
e) In the United Kingdom, mpg uses the imperial gallon, $4.54609$ L. Write the conversion $u$ from miles per imperial gallon to miles per US gallon, then find the US rating and the consumption in L per $100$ km of a car rated $36$ mpg (imperial).

Show the solution

Answers

a) $k(30) \approx 12.75$ km/L; $h(k(30)) \approx 7.84$ L per $100$ km
b) $c(m) = \frac{235.21}{m}$ , $m > 0$ ; $k(h(30))$ has no meaning (its units are not those of $k$ 's input)
c) A saves $\approx 3.92$ L, B $\approx 1.18$ L per $100$ km: A saves more
d) $120$ mpg
e) $u(i) = \frac{3.785412}{4.54609}\,i \approx 0.8327\,i$ ; $u(36) \approx 29.98$ US mpg; $\approx 7.85$ L per $100$ km

a)

Follow the units: $m$ miles per gallon is $m \times 1.609344$ km per $3.785412$ L, so per litre, $k(m) = \frac{1.609344}{3.785412}\,m \approx 0.42514\,m$ km/L. Then $k(30) \approx 12.75$ km/L and $h(12.754) = \frac{100}{12.754} \approx 7.84$ L per $100$ km. Writing the units at every step is the check: multiplying by $3.785412$ instead of dividing would give km times litres per gallon squared, a unit that means nothing, and $0.4251$ km/L for $1$ mpg is sensible because a mile is less than half of a gallon's worth of litres.

b)

Compose, inner function first: $c(m) = h(k(m)) = \frac{100}{0.425144\,m} = \frac{100 \times 3.785412}{1.609344}\cdot\frac{1}{m} \approx \frac{235.21}{m}$ . Domain of the formula: $m \ne 0$ ; domain of the MODEL: $m > 0$ , a fuel economy being positive. The composition in the other order, $k(h(30))$ , computes $h(30) = \frac{10}{3}$ and feeds it to $k$ as if it were in mpg: $k$ expects miles per US gallon, and $h(30)$ would be litres per $100$ km of a car doing $30$ km per litre, another unit. A composition makes sense only when the OUTPUT of the inner function has the units of the INPUT of the outer one. That is the real content of the domain condition in a model.

c)

Car A: $c(15) - c(20) = 235.21\left(\frac{1}{15} - \frac{1}{20}\right) = \frac{235.21}{60} \approx 3.92$ L per $100$ km. Car B: $c(40) - c(50) = 235.21\left(\frac{1}{40} - \frac{1}{50}\right) = \frac{235.21}{200} \approx 1.18$ L per $100$ km. The SMALLER gain in mpg, $5$ against $10$ , saves more than three times as much fuel. The figure shows why: $c$ is a multiple of $\frac{1}{m}$ , whose graph is steep for small $m$ and flat for large $m$ . The common denominator is the algebra step: $\frac{1}{15} - \frac{1}{20} = \frac{4 - 3}{60}$ .

d)

We need $235.21\left(\frac{1}{40} - \frac{1}{m}\right) = \frac{235.21}{60}$ . The constant CANCELS, which is why the answer is exact: $\frac{1}{m} = \frac{1}{40} - \frac{1}{60} = \frac{3 - 2}{120} = \frac{1}{120}$ , so $m = 120$ mpg. Tripling the rating of car B saves as much as the $5$ mpg of car A. Check: $\frac{235.21}{40} - \frac{235.21}{120} = 5.880 - 1.960 = 3.92$ . Inverting $\frac{1}{40} - \frac{1}{60}$ as $40 - 60$ is the slip; subtract the fractions first, invert last.

e)

$i$ miles per imperial gallon is $i$ miles per $4.54609$ L, so per US gallon of $3.785412$ L it is $u(i) = \frac{3.785412}{4.54609}\,i \approx 0.8327\,i$ miles. So $u(36) \approx 29.98$ mpg (US), and the consumption is $c(u(36)) = \frac{235.21}{29.976} \approx 7.85$ L per $100$ km. The same car is rated $36$ in the United Kingdom and about $30$ in the United States, only because the gallons differ; the consumption in L per $100$ km, which does not depend on gallons, is the rating to compare. As a composition, $c \circ u$ converts in two layers, and each layer is checked by its units.

Corrected exercises: functions, trigonometric and exponential functions (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: Sums, quotients and compositions: the domain comes from the construction

Exercise 2: Shifting and scaling: factor the inside, then keep the order

Exercise 3: Radians, the unit circle and the six trigonometric functions

Exercise 4: Identities and trigonometric equations on an interval: factor, never divide

Exercise 5: The general sine function: amplitude, period and shift, from a formula and from a graph

Part B: problems and reasoning (/50)

Exercise 6: Exponential functions and the laws of exponents: rewrite as one power

Exercise 7: Period, range and domain of combined trigonometric and exponential functions

Exercise 8: Five statements to correct

Exercise 9: A tide modelled by a sinusoid: when can the ship leave?

Exercise 10: A final exam problem: fuel economy, a composition of unit conversions

See also

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