Exercise 0.1: Find the domain: twelve direct calculations

For each function, find the domain and write it in interval notation. When the domain is the union of two intervals, give the left interval then the right one; when it is $\mathbb{R}$ minus a few values, give those excluded values.

One method for all twelve lines: spot each of the three restrictions present (a zero denominator, an EVEN root of a negative number, a logarithm of a number that is not strictly positive), write the condition each one imposes, solve it, then take the intersection of all the conditions. A root in the denominator makes the inequality strict; a cube root restricts nothing.

a) $f(x)=4x^{3}-x+7$
b) $f(x)=\sqrt{10-4x}$
c) $f(x)=\frac{2x-1}{x^{2}+x-12}$
d) $f(x)=\frac{x}{|x|-3}$
e) $f(x)=\sqrt[3]{\frac{x+2}{x-7}}$
f) $f(x)=\sqrt{x^{2}-2x-8}$
g) $f(x)=\frac{3}{\sqrt{12+x-x^{2}}}$
h) $f(x)=\frac{\sqrt{x+5}}{x-1}$
i) $f(x)=\ln\left(x^{2}+3x-10\right)$
j) $f(x)=\frac{1}{\ln(x+2)}$
k) $f(x)=\frac{1}{e^{2x}-4e^{x}+3}$
l) $f(x)=\arcsin(3-2x)$

Type your answers, the page tells you right or wrong 0/19

a)

domain ,

b)

domain ,

c)

excluded value, the smaller

excluded value, the larger

d)

excluded value, the smaller

excluded value, the larger

e)

only excluded value

f)

left interval ,

right interval ,

g)

domain ,

h)

left interval ,

right interval ,

i)

left interval ,

right interval ,

j)

left interval ,

right interval ,

k)

excluded value, the smaller

excluded value, the larger

l)

domain ,

Show the solution

Answers

a) $\left(-\infty, +\infty\right)$
b) $\left(-\infty, \frac{5}{2}\right]$
c) $\mathbb{R}\setminus\{-4, 3\}$
d) $\mathbb{R}\setminus\{-3, 3\}$
e) $\mathbb{R}\setminus\{7\}$
f) $\left(-\infty, -2\right]\cup\left[4, +\infty\right)$
g) $\left(-3, 4\right)$
h) $\left[-5, 1\right)\cup\left(1, +\infty\right)$
i) $\left(-\infty, -5\right)\cup\left(2, +\infty\right)$
j) $\left(-2, -1\right)\cup\left(-1, +\infty\right)$
k) $\mathbb{R}\setminus\{0, \ln 3\}$ , $\ln 3\approx 1.099$
l) $[1, 2]$

The move that decides everything: write the CONDITION before solving it, one per restriction, and never forget one. $\frac{\sqrt{x+5}}{x-1}$ carries two, $x+5\ge 0$ and $x-1\neq 0$ ; so does $\frac{1}{\ln(x+2)}$ , $x+2>0$ and $\ln(x+2)\neq 0$ . The domain is what satisfies them ALL.

a)

A polynomial has no restriction: $\text{dom}\,f=\mathbb{R}=\left(-\infty, +\infty\right)$ .

b)

$10-4x\ge 0\iff -4x\ge-10\iff x\le\frac{5}{2}$ , the inequality flips because we divide by $-4$ . $\text{dom}\,f=\left(-\infty, \frac{5}{2}\right]$ , endpoint included since $\sqrt{0}$ exists. Pitfall: keeping the direction and writing $x\ge\frac{5}{2}$ .

c)

$x^{2}+x-12=(x+4)(x-3)\neq 0$ : $x\neq-4$ and $x\neq 3$ . $\text{dom}\,f=\mathbb{R}\setminus\{-4, 3\}$ . The numerator plays no role.

d)

$|x|-3\neq 0\iff|x|\neq 3\iff x\neq\pm 3$ . $\text{dom}\,f=\mathbb{R}\setminus\{-3, 3\}$ . Pitfall: excluding only $x=3$ , forgetting that $|-3|=3$ as well.

e)

A CUBE root accepts negatives: only the denominator matters, $x\neq 7$ . $\text{dom}\,f=\mathbb{R}\setminus\{7\}$ . Pitfall: imposing $\frac{x+2}{x-7}\ge 0$ as for a square root, which would wrongly remove the whole interval $\left(-2, 7\right)$ .

f)

$x^{2}-2x-8=(x+2)(x-4)\ge 0$ : the quadratic is positive outside its roots. $\text{dom}\,f=\left(-\infty, -2\right]\cup\left[4, +\infty\right)$ , endpoints included.

g)

The root is in the denominator: $12+x-x^{2}>0$ , STRICTLY. Now $12+x-x^{2}=-(x-4)(x+3)$ , positive BETWEEN the roots. $\text{dom}\,f=\left(-3, 4\right)$ . Pitfall: using square brackets, when at $-3$ and at $4$ we would divide by $0$ .

h)

Two conditions: $x+5\ge 0$ , that is $x\ge-5$ , and $x-1\neq 0$ , that is $x\neq 1$ . $\text{dom}\,f=\left[-5, 1\right)\cup\left(1, +\infty\right)$ . Pitfall: forgetting to remove $1$ , which does belong to the domain of the root.

i)

$x^{2}+3x-10=(x+5)(x-2)>0$ , strictly, outside the roots. $\text{dom}\,f=\left(-\infty, -5\right)\cup\left(2, +\infty\right)$ . Pitfall: using square brackets, when $\ln 0$ does not exist.

j)

Two conditions: $x+2>0$ , that is $x>-2$ , and $\ln(x+2)\neq 0$ , that is $x+2\neq 1$ , so $x\neq-1$ . $\text{dom}\,f=\left(-2, -1\right)\cup\left(-1, +\infty\right)$ . Pitfall: writing $\ln(x+2)\neq 0\iff x\neq-2$ , confusing the zero of the logarithm ( $u=1$ ) with its boundary ( $u=0$ ).

k)

Let $u=e^{x}$ : $u^{2}-4u+3=(u-1)(u-3)\neq 0$ , so $e^{x}\neq 1$ and $e^{x}\neq 3$ , that is $x\neq 0$ and $x\neq\ln 3\approx 1.099$ . $\text{dom}\,f=\mathbb{R}\setminus\{0, \ln 3\}$ . Pitfall: answering $x\neq 1$ and $x\neq 3$ , which are values of $e^{x}$ , not of $x$ .

l)

$\arcsin$ only accepts $[-1, 1]$ : $-1\le 3-2x\le 1\iff-4\le-2x\le-2\iff 2\ge x\ge 1$ . $\text{dom}\,f=[1, 2]$ . Pitfall: dividing by $-2$ without reversing both inequalities, which gives the empty interval $[2, 1]$ .

Check: evaluate $f$ on the calculator at a point just inside and just outside each endpoint, for example at $3.999$ and at $4.001$ on line g); the calculator must answer “error” exactly on the excluded side, and at the excluded values themselves.

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Exercise 0.2: Compose two functions and invert: ten direct calculations

For each line, compute the value asked for: a composition at a point, $(f\circ g)(a)=f\left(g(a)\right)$ , or the value of an inverse at a point, $f^{-1}(b)$ . When a value does not exist, say so.

A composition is computed from the inside out: first evaluate the function written to the RIGHT of the circle, then feed the result to the one on the left. For $f^{-1}(b)$ , there is no need to write the formula of the inverse: look for the number $x$ such that $f(x)=b$ , then check that it belongs to the domain of $f$ , or of its restriction. If no $x$ works, $b$ is not in the range of $f$ and $f^{-1}(b)$ does not exist.

a) $f(x)=3x-1$ and $g(x)=x^{2}+2$ : $(f\circ g)(2)$ , then $(g\circ f)(2)$ .
b) $f(x)=5-2x$ : $(f\circ f)(3)$ .
c) $f(x)=\sqrt{x}$ and $g(x)=x^{2}-7$ : $(f\circ g)(4)$ , then $(g\circ f)(4)$ .
d) $f(x)=\frac{1}{x}$ and $g(x)=2x+3$ : $(f\circ g)\left(\frac{1}{2}\right)$ , then $(g\circ f)\left(\frac{1}{2}\right)$ .
e) $f(x)=|x-4|$ and $g(x)=x^{2}-3x$ : $(f\circ g)(1)$ , then $(g\circ f)(1)$ .
f) $f(x)=4x+7$ : $f^{-1}(-5)$ .
g) $f(x)=\frac{3x+2}{x-1}$ : $f^{-1}(5)$ .
h) $f(x)=\sqrt{2x+5}$ : $f^{-1}(3)$ , then $f^{-1}(-1)$ .
i) $f(x)=x^{2}-6$ restricted to $\left[0, +\infty\right)$ : $f^{-1}(10)$ .
j) $f(x)=3\cdot 2^{x}$ : $f^{-1}(24)$ ; then $g(x)=\ln(x-2)$ : $g^{-1}(1)$ .

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

a)

(f\circ g)(2)

(g\circ f)(2)

b)

(f\circ f)(3)

c)

(f\circ g)(4)

(g\circ f)(4)

d)

(f\circ g)\left(\frac{1}{2}\right)

(g\circ f)\left(\frac{1}{2}\right)

e)

(f\circ g)(1)

(g\circ f)(1)

f)

f^{-1}(-5)

g)

f^{-1}(5)

h)

f^{-1}(3)

f^{-1}(-1)

i)

f^{-1}(10)

j)

f^{-1}(24)

g^{-1}(1)

Show the solution

Answers

a) $(f\circ g)(2)=17$ ; $(g\circ f)(2)=27$
b) $(f\circ f)(3)=f(-1)=7$
c) $(f\circ g)(4)=3$ ; $(g\circ f)(4)=-3$
d) $(f\circ g)\left(\frac{1}{2}\right)=\frac{1}{4}$ ; $(g\circ f)\left(\frac{1}{2}\right)=7$
e) $(f\circ g)(1)=6$ ; $(g\circ f)(1)=0$
f) $f^{-1}(-5)=-3$
g) $f^{-1}(5)=\frac{7}{2}$
h) $f^{-1}(3)=2$ ; $f^{-1}(-1)$ does not exist
i) $f^{-1}(10)=4$
j) $f^{-1}(24)=3$ ; $g^{-1}(1)=2+e\approx 4.718$

The move that decides everything: read $(f\circ g)(a)$ from RIGHT to LEFT, $g$ first, and read $f^{-1}(b)$ as an EQUATION, $f(x)=b$ , never as a reciprocal $\frac{1}{f(b)}$ . The two traps of the chapter fit in these two readings.

a)

$(f\circ g)(2)=f\left(g(2)\right)=f(6)=17$ ; $(g\circ f)(2)=g\left(f(2)\right)=g(5)=27$ . The two results differ: composition is not commutative. Pitfall: computing $g$ last for $(f\circ g)$ .

b)

$(f\circ f)(3)=f\left(f(3)\right)=f(-1)=5+2=7$ . Pitfall: $(f\circ f)(3)$ is not $\left(f(3)\right)^{2}=1$ ; we compose, we do not multiply.

c)

$(f\circ g)(4)=f(16-7)=f(9)=3$ ; $(g\circ f)(4)=g(2)=4-7=-3$ .

d)

$(f\circ g)\left(\frac{1}{2}\right)=f(1+3)=f(4)=\frac{1}{4}$ ; $(g\circ f)\left(\frac{1}{2}\right)=g(2)=7$ .

e)

$(f\circ g)(1)=f(1-3)=f(-2)=|-6|=6$ ; $(g\circ f)(1)=g\left(|1-4|\right)=g(3)=9-9=0$ . Pitfall: dropping the absolute value and answering $-6$ .

f)

$4x+7=-5\iff 4x=-12\iff x=-3$ , so $f^{-1}(-5)=-3$ . Check: $f(-3)=-12+7=-5$ . Pitfall: $f^{-1}(-5)=\frac{1}{f(-5)}=-\frac{1}{13}$ , which confuses the inverse function with the reciprocal.

g)

$\frac{3x+2}{x-1}=5\iff 3x+2=5x-5\iff 7=2x\iff x=\frac{7}{2}$ , and $\frac{7}{2}\neq 1$ is in the domain. $f^{-1}(5)=\frac{7}{2}$ . Check: $f\left(\frac{7}{2}\right)=\frac{12.5}{2.5}=5$ . Pitfall: computing $f(5)=\frac{17}{4}$ .

h)

$\sqrt{2x+5}=3\iff 2x+5=9\iff x=2$ , so $f^{-1}(3)=2$ . On the other hand, a square root is never negative: the range of $f$ is $\left[0, +\infty\right)$ , $-1$ is not in it, and $f^{-1}(-1)$ does not exist. Pitfall: squaring, $2x+5=1$ , and answering $x=-2$ , when $f(-2)=1\neq-1$ .

i)

$x^{2}-6=10\iff x^{2}=16\iff x=\pm 4$ , but the restriction requires $x\ge 0$ : $f^{-1}(10)=4$ . Pitfall: answering $-4$ , or both; an inverse gives ONE value, taken in the domain of the restriction.

j)

$3\cdot 2^{x}=24\iff 2^{x}=8\iff x=3$ , so $f^{-1}(24)=3$ . Then $\ln(x-2)=1\iff x-2=e\iff x=2+e\approx 4.718$ , so $g^{-1}(1)=2+e$ . Pitfall: writing $x-2=1$ , when $\ln(x-2)=1$ means $x-2=e^{1}$ .

Check: for an inverse, substitute the value found into $f$ and you must get $b$ back; for a composition, redo the calculation by first writing the formula of $f\circ g$ , for example $(f\circ g)(x)=3(x^{2}+2)-1=3x^{2}+5$ on line a), which does give $17$ at $x=2$ .

Exercise 1: The domain: the only three restrictions

Find the domain of each of these functions, and write it in interval notation.

a) $f(x)=\frac{x+3}{x^{2}-4}$
b) $g(x)=\sqrt{6-2x}$
c) $h(x)=\frac{\sqrt{x^{2}-x-6}}{x-4}$
d) $k(x)=\frac{1}{\sqrt{9-x^{2}}}$ . Compare its domain with that of $\sqrt{9-x^{2}}$ and explain the difference.

Show the solution

Answers

a) $\mathbb{R}\setminus\{-2, 2\}$
b) $\left(-\infty, 3\right]$
c) $\left(-\infty, -2\right]\cup\left[3, 4\right)\cup\left(4, +\infty\right)$
d) $\left(-3, 3\right)$ versus $\left[-3, 3\right]$ : root in the denominator

a)

A quotient forbids a zero denominator, and that is the only restriction here. $x^{2}-4=(x-2)(x+2)$ is zero at $x=2$ and $x=-2$ . So $\text{dom}\,f=\left(-\infty, -2\right)\cup\left(-2, 2\right)\cup\left(2, +\infty\right)$ , which can also be written $\mathbb{R}\setminus\{-2, 2\}$ . The numerator plays NO role in the domain: a solution that excludes $x=-3$ because the numerator vanishes there confuses a zero of the function with a forbidden value.

b)

A root with an EVEN index requires a radicand greater than or equal to zero: $6-2x\ge 0$ , that is $x\le 3$ . So $\text{dom}\,g=\left(-\infty, 3\right]$ . The bracket is square at $3$ because $\sqrt{0}=0$ exists: that is exactly the difference with part d).

c)

Two restrictions add up. First the root: $x^{2}-x-6=(x-3)(x+2)\ge 0$ . The sign chart of the quadratic, positive outside its roots, gives $x\le-2$ or $x\ge 3$ . Then the denominator: $x\neq 4$ . Combining the two conditions, $\text{dom}\,h=\left(-\infty, -2\right]\cup\left[3, 4\right)\cup\left(4, +\infty\right)$ , with a parenthesis that is always used on the side of infinity. Quick check: $h(0)$ requires $\sqrt{-6}$ , so $0$ must be outside the domain ✓, and $h(5)=\frac{\sqrt{14}}{1}$ exists ✓.

d)

Here the root is in the DENOMINATOR, so it must be defined AND nonzero: $9-x^{2}>0$ , a STRICT inequality, that is $-3<x<3$ and $\text{dom}\,k=\left(-3, 3\right)$ . For $\sqrt{9-x^{2}}$ , the inequality is not strict and the domain is $\left[-3, 3\right]$ . The two endpoints $\pm 3$ belong to the second domain and not to the first, because there the radicand is $0$ , a perfectly acceptable value under a root but forbidden in a denominator. This is the most profitable distinction of the chapter: it decides the domain, hence the vertical asymptotes, hence the limits you will have to compute in the next chapter.

Exercise 2: Absolute value and piecewise-defined function

Let $f(x)=|x-1|+|x+2|$ .

a) Write $f$ without absolute values, as a piecewise-defined function.
b) Sketch the graph of $f$ and give its range.
c) Solve $f(x)=7$ .
d) Interpret $f(x)$ as a sum of distances, and explain where the flat part comes from.

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Answers

a) $-2x-1$ if $x<-2$ ; $3$ if $-2\le x<1$ ; $2x+1$ if $x\ge 1$
b) Range $\left[3, +\infty\right)$
c) $\{-4, 3\}$
d) Sum of the distances to $1$ and to $-2$ : flat part at height $3$

a)

The two absolute values change definition at $x=1$ and at $x=-2$ : these are the breakpoints, and they split $\mathbb{R}$ into three pieces. If $x<-2$ , then $x-1<0$ and $x+2<0$ , so $f(x)=-(x-1)-(x+2)=-2x-1$ . If $-2\le x<1$ , then $x-1<0$ but $x+2\ge 0$ , so $f(x)=-(x-1)+(x+2)=3$ . If $x\ge 1$ , both are positive and $f(x)=(x-1)+(x+2)=2x+1$ . The move to remember: do not guess the signs, place the two breakpoints on a number line and read each interval.

b)

The graph is made of three pieces: a ray of slope $-2$ up to $(-2, 3)$ , a horizontal segment at height $3$ between $x=-2$ and $x=1$ , then a ray of slope $2$ starting from $(1, 3)$ . The pieces join continuously: $-2(-2)-1=3$ and $2(1)+1=3$ . The range is $\left[3, +\infty\right)$ : the value $3$ is reached on a whole segment, not at a single point, and nothing below it is reached.

c)

Solve piece by piece, checking each time that the solution really belongs to the piece. On $x<-2$ : $-2x-1=7$ gives $x=-4$ , which does satisfy $-4<-2$ ✓. On $-2\le x<1$ : $3=7$ is impossible, no solution. On $x\ge 1$ : $2x+1=7$ gives $x=3$ , which satisfies $3\ge 1$ ✓. The solution set is $\{-4, 3\}$ . Check: $f(-4)=|-5|+|-2|=5+2=7$ ✓ and $f(3)=|2|+|5|=2+5=7$ ✓. Forgetting to check that a solution belongs to its piece is the classic error: it produces extraneous solutions that cost a mark each.

d)

$|x-1|$ is the distance from $x$ to the point $1$ , and $|x+2|=|x-(-2)|$ is the distance from $x$ to the point $-2$ . So $f(x)$ is the sum of the distances from $x$ to two points that are $3$ apart. When $x$ is BETWEEN them, this sum equals exactly the gap, that is $3$ , whatever the position: that is the flat part. When $x$ leaves the interval, each extra step moves it away from TWO points at once, hence the slopes $\pm 2$ . This reading gives the graph and the minimum without any calculation, and it generalizes: $|x-a|+|x-b|$ has minimum $|b-a|$ , reached on the whole segment $[a, b]$ .

Exercise 3: Operations on functions: the domain does not simply carry over

Let $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{4-x}$ .

a) Give $\text{dom}\,f$ and $\text{dom}\,g$ .
b) Give $\text{dom}(f+g)$ , $\text{dom}(f\cdot g)$ and $\text{dom}\left(\frac{f}{g}\right)$ , and state precisely what sets the third apart from the first two.
c) Compute $(f+g)(0)$ and $(f+g)(3)$ .
d) Show that $(f\cdot g)(x)=\sqrt{(x+1)(4-x)}$ and find its maximum without differentiating.

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Answers

a) $\left[-1, +\infty\right)$ and $\left(-\infty, 4\right]$
b) $\left[-1, 4\right]$ , $\left[-1, 4\right]$ and $\left[-1, 4\right)$
c) $(f+g)(0)=(f+g)(3)=3$
d) Maximum $\frac{5}{2}$ at $x=\frac{3}{2}$

a)

$\text{dom}\,f=\left[-1, +\infty\right)$ because $x+1\ge 0$ , and $\text{dom}\,g=\left(-\infty, 4\right]$ because $4-x\ge 0$ .

b)

The general rule: the domain of a sum, a difference or a product is the INTERSECTION of the domains, because both functions must exist. Here $\text{dom}(f+g)=\text{dom}(f\cdot g)=\left[-1, +\infty\right)\cap\left(-\infty, 4\right]=\left[-1, 4\right]$ . For the quotient one condition is added, and only one: the denominator must not be zero. Now $g(x)=0$ for $x=4$ . So $\text{dom}\left(\frac{f}{g}\right)=\left[-1, 4\right)$ . This is the point that regularly costs a mark on the exam: remove the zeros of the DENOMINATOR, never those of the numerator, and remove them from an intersection that has already been computed.

c)

$(f+g)(0)=\sqrt{1}+\sqrt{4}=1+2=3$ . $(f+g)(3)=\sqrt{4}+\sqrt{1}=2+1=3$ as well, which is no coincidence: the sum is symmetric about the midpoint of $\left[-1, 4\right]$ , that is $x=1.5$ , since swapping $x+1$ and $4-x$ amounts to reflecting about that point.

d)

$(f\cdot g)(x)=\sqrt{x+1}\cdot\sqrt{4-x}=\sqrt{(x+1)(4-x)}$ : the product of two roots is the root of the product, provided both radicands are nonnegative, which is exactly the domain found in part b). The radicand $(x+1)(4-x)=-x^{2}+3x+4$ is a quadratic with a negative leading coefficient, so a parabola opening downward, whose vertex is at $x=-\frac{3}{2(-1)}=\frac{3}{2}$ . Since the square root is increasing, it reaches its maximum at the same place as its radicand: $(f\cdot g)\left(\frac{3}{2}\right)=\sqrt{\left(\frac{5}{2}\right)\left(\frac{5}{2}\right)}=\frac{5}{2}$ . No derivative was needed, and the argument “the square root is increasing, so it preserves extrema” is used throughout the course.

Exercise 4: Composition, and the domain that is not the one you expect

Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$ .

a) Write $(f\circ g)(x)$ and give its domain.
b) Write $(g\circ f)(x)$ and give its domain.
c) Are these two functions equal? Support your answer with a numerical counterexample.
d) Decompose $h(x)=\frac{1}{\sqrt{x^{2}+3}}$ into a composition of three simple functions.

Show the solution

Answers

a) $\frac{1}{\sqrt{x}-2}$ , domain $\left[0, 4\right)\cup\left(4, +\infty\right)$
b) $\sqrt{\frac{1}{x-2}}$ , domain $\left(2, +\infty\right)$
c) No: $-3.732$ versus $1$ at $x=3$
d) $h=w\circ v\circ u$ with $u=x^{2}+3$ , $v=\sqrt{t}$ , $w=\frac{1}{s}$

a)

$(f\circ g)(x)=f\left(\sqrt{x}\right)=\frac{1}{\sqrt{x}-2}$ . The domain is built in TWO steps, never in one. First step: $x$ must belong to the domain of $g$ , so $x\ge 0$ . Second step: $g(x)$ must belong to the domain of $f$ , so $\sqrt{x}\neq 2$ , that is $x\neq 4$ . Conclusion: $\text{dom}(f\circ g)=\left[0, 4\right)\cup\left(4, +\infty\right)$ . Reading the domain off the simplified expression $\frac{1}{\sqrt{x}-2}$ would give the right result here by luck, but the two-step method is the only safe one.

b)

$(g\circ f)(x)=g\left(\frac{1}{x-2}\right)=\sqrt{\frac{1}{x-2}}$ . First step: $x\neq 2$ . Second step: $\frac{1}{x-2}\ge 0$ , which requires $x-2>0$ , that is $x>2$ . Conclusion: $\text{dom}(g\circ f)=\left(2, +\infty\right)$ . The two domains are therefore radically different, which is already enough to settle part c).

c)

No. Two functions are equal when they have the same domain AND the same value at every point, and here the first condition already fails: $0$ belongs to $\text{dom}(f\circ g)$ and not to $\text{dom}(g\circ f)$ . A numerical counterexample at a point common to both domains, $x=3$ : $(f\circ g)(3)=\frac{1}{\sqrt{3}-2}\approx-3.732$ , while $(g\circ f)(3)=\sqrt{\frac{1}{1}}=1$ . Composition is NOT commutative, and the reading order is from the inside out: in $f\circ g$ , it is $g$ that acts first.

d)

Read $h$ from the inside out, which is exactly the order in which a calculator would evaluate it. Square and add $3$ : $u(x)=x^{2}+3$ . Take the root: $v(t)=\sqrt{t}$ . Take the reciprocal: $w(s)=\frac{1}{s}$ . Then $h=w\circ v\circ u$ . Check at $x=1$ : $u(1)=4$ , $v(4)=2$ , $w(2)=\frac{1}{2}$ , and $h(1)=\frac{1}{\sqrt{4}}=\frac{1}{2}$ ✓. This decomposition is not a pointless exercise: it is exactly the order in which the chain rule will differentiate $h$ in the chapter on the derivative, one layer after another, starting from the outside.

Exercise 5: Reading a transformed graph

The figure shows the parabola $y=x^{2}$ and the graph of a function $g$ obtained from it by transformations.

a) Identify the transformations and give the expression of $g$ .
b) In what order must the transformations of $y=-2f(x-1)+3$ be applied? Justify that the order matters.
c) If $(a, b)$ is a point on the graph of $f$ , which point on the graph of $y=-2f(x-1)+3$ corresponds to it?
d) $f$ has domain $\left[-2, 6\right]$ and range $\left[0, 5\right]$ . Give the domain and the range of $y=-2f(x-1)+3$ .

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Answers

a) $g(x)=(x-1)^{2}-4$
b) Horizontal translation, stretch, reflection, vertical translation
c) $(a+1, -2b+3)$
d) Domain $\left[-1, 7\right]$ , range $\left[-7, 3\right]$

a)

The vertex of $y=x^{2}$ is at $(0, 0)$ ; that of the second curve is read at $(1, -4)$ . The two curves have the same opening, so there is no vertical stretch: it is a translation of $1$ to the right and $4$ down, that is $g(x)=(x-1)^{2}-4$ . Check on a second point: $g(3)=4-4=0$ ✓, and the curve does cross the $x$ -axis at $x=3$ , as well as at $x=-1$ .

b)

The order is that of the operations on $x$ , then on $y$ : first the horizontal translation $x\mapsto x-1$ , then the vertical stretch by a factor of $2$ , then the reflection in the $x$ -axis, and finally the vertical translation of $+3$ . The order matters because these transformations do not commute: reflecting then moving up $3$ gives $-f(x)+3$ , while moving up $3$ then reflecting gives $-\left(f(x)+3\right)=-f(x)-3$ , two curves $6$ units apart. The reading rule that never fails: what is ATTACHED to $x$ acts the opposite way and first, what is outside acts the natural way and last.

c)

$x-1=a$ gives $x=a+1$ , and the $y$ -coordinate becomes $-2b+3$ . The image point is therefore $(a+1, \,-2b+3)$ . The shift goes in the direction OPPOSITE to the one written in the parentheses, which is the most frequent source of error: $f(x-1)$ shifts to the RIGHT.

d)

Domain: each $x$ -coordinate is increased by $1$ , so $\left[-2+1, 6+1\right]=\left[-1, 7\right]$ . Range: apply $b\mapsto-2b+3$ to the values of $\left[0, 5\right]$ . Since the function $b\mapsto-2b+3$ is DECREASING, it swaps the endpoints: $b=0$ gives $3$ and $b=5$ gives $-7$ . The range is therefore $\left[-7, 3\right]$ , and not $\left[3, -7\right]$ , which means nothing. Writing the range interval in the wrong order, or forgetting the reversal caused by the negative coefficient, costs the whole mark for the question.

Exercise 6: Inverse functions: existence, domain and range

Let $f(x)=\frac{2x-1}{x+3}$ .

a) Give the domain and the range of $f$ .
b) Find $f^{-1}$ and give its domain and its range.
c) Check that $f\left(f^{-1}(x)\right)=x$ at a point of your choice, then in general.
d) Let $g(x)=x^{2}-4x$ . Why does $g$ have no inverse on $\mathbb{R}$ ? Give the inverse of its restriction to $\left[2, +\infty\right)$ , then that of its restriction to $\left(-\infty, 2\right]$ .

Show the solution

Answers

a) Domain $\mathbb{R}\setminus\{-3\}$ , range $\mathbb{R}\setminus\{2\}$
b) $f^{-1}(x)=\frac{3x+1}{2-x}$ , domain $\mathbb{R}\setminus\{2\}$ , range $\mathbb{R}\setminus\{-3\}$
c) $f^{-1}(5)=-\frac{16}{3}$ and $f\left(f^{-1}(x)\right)=x$
d) Not one-to-one; $2+\sqrt{x+4}$ and $2-\sqrt{x+4}$ on $\left[-4, +\infty\right)$

a)

$\text{dom}\,f=\mathbb{R}\setminus\{-3\}$ . For the range, look for the values $y$ that are reached: $y=\frac{2x-1}{x+3}$ is equivalent to $y(x+3)=2x-1$ , that is $x(y-2)=-1-3y$ . This equation has a solution in $x$ for every $y\neq 2$ , and none for $y=2$ , since it would become $0=-7$ . So the range is $\mathbb{R}\setminus\{2\}$ . We recognize $2$ as the ratio of the coefficients of $x$ , which is the horizontal asymptote of this rational function.

b)

From $x(y-2)=-1-3y$ we get $x=\frac{-1-3y}{y-2}=\frac{3y+1}{2-y}$ . So $f^{-1}(x)=\frac{3x+1}{2-x}$ . Its domain is the range of $f$ , that is $\mathbb{R}\setminus\{2\}$ , and its range is the domain of $f$ , that is $\mathbb{R}\setminus\{-3\}$ . This swap is not a coincidence to be checked case by case: it is the very definition of an inverse, and stating it earns the justification mark.

c)

At $x=5$ : $f^{-1}(5)=\frac{16}{-3}=-\frac{16}{3}$ , then $f\left(-\frac{16}{3}\right)=\frac{-\frac{32}{3}-1}{-\frac{16}{3}+3}=\frac{-\frac{35}{3}}{-\frac{7}{3}}=5$ ✓. In general: $f\left(\frac{3x+1}{2-x}\right)=\frac{2\cdot\frac{3x+1}{2-x}-1}{\frac{3x+1}{2-x}+3}=\frac{\frac{6x+2-(2-x)}{2-x}}{\frac{3x+1+3(2-x)}{2-x}}=\frac{7x}{7}=x$ ✓. Multiplying top and bottom by $(2-x)$ clears the stacked fractions in one line: that is the manipulation to know, and it will come back with every rational function of this type.

d)

$g$ is not one-to-one on $\mathbb{R}$ : $g(0)=0$ and $g(4)=0$ , two inputs for the same output, so no function can send $0$ back to both at once. Graphically, a horizontal line cuts the parabola at two points. Completing the square, $g(x)=(x-2)^{2}-4$ , which gives the two natural restrictions on either side of the vertex $x=2$ . On $\left[2, +\infty\right)$ : $y=(x-2)^{2}-4$ gives $x-2=\sqrt{y+4}$ , the POSITIVE root since $x\ge 2$ , hence $g^{-1}(x)=2+\sqrt{x+4}$ , with domain $\left[-4, +\infty\right)$ . On $\left(-\infty, 2\right]$ : $x-2=-\sqrt{y+4}$ , hence $g^{-1}(x)=2-\sqrt{x+4}$ , with the same domain. The sign in front of the root is ALWAYS dictated by the chosen restriction, never by habit.

Exercise 7: The unit circle and periodic functions

All angle measures are in radians.

a) Give the exact values of $\sin\frac{5\pi}{6}$ , $\cos\frac{5\pi}{6}$ , $\tan\frac{7\pi}{4}$ and $\cos\frac{4\pi}{3}$ .
b) Solve $2\cos x=\sqrt{3}$ on $\left[0, 2\pi\right)$ .
c) For $h(x)=3\sin\left(2x-\frac{\pi}{3}\right)+1$ , give the amplitude, the period, the phase shift and the range.
d) Give the domain of $\tan$ and that of $\sec$ , and say what these domains have in common.

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

a)

\sin\frac{5\pi}{6}

\cos\frac{5\pi}{6}

\tan\frac{7\pi}{4}

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

b)

Smaller solution (in multiples of π)

Larger solution (in multiples of π)

c)

Amplitude

Period (in multiples of π)

Phase shift (in multiples of π)

Range ,

d)

\tan

does not exist at

The domain of

\sec

is

Show the solution

Answers

a) $\frac{1}{2}$ ; $-\frac{\sqrt{3}}{2}$ ; $-1$ ; $-\frac{1}{2}$
b) $\frac{\pi}{6}$ and $\frac{11\pi}{6}$
c) Amplitude 3, period $\pi$ , phase shift $\frac{\pi}{6}$ , range $\left[-2, 4\right]$
d) $x\neq\frac{\pi}{2}+k\pi$ for both

a)

Bring each angle back to its reference angle and read the sign from the quadrant. $\frac{5\pi}{6}$ is in the second quadrant, with reference angle $\frac{\pi}{6}$ : sine is positive there and cosine negative, so $\sin\frac{5\pi}{6}=\frac{1}{2}$ and $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$ . $\frac{7\pi}{4}$ is in the fourth quadrant, with reference angle $\frac{\pi}{4}$ : tangent is negative there, so $\tan\frac{7\pi}{4}=-1$ . $\frac{4\pi}{3}$ is in the third quadrant, with reference angle $\frac{\pi}{3}$ : cosine is negative there, so $\cos\frac{4\pi}{3}=-\frac{1}{2}$ . The method “reference angle, then sign from the quadrant” beats memorizing a table of twenty-four values.

b)

$2\cos x=\sqrt{3}$ gives $\cos x=\frac{\sqrt{3}}{2}$ . The reference angle is $\frac{\pi}{6}$ , and cosine is positive in the first and fourth quadrants. On $\left[0, 2\pi\right)$ , the solutions are therefore $x=\frac{\pi}{6}$ and $x=2\pi-\frac{\pi}{6}=\frac{11\pi}{6}$ . Giving only one solution is the most frequent error and costs half the question: a trigonometric equation always has two families of solutions per turn, except at the extreme values $\pm 1$ .

c)

Read it in the form $a\sin\left(b(x-c)\right)+d$ . Here $3\sin\left(2x-\frac{\pi}{3}\right)+1=3\sin\left(2\left(x-\frac{\pi}{6}\right)\right)+1$ : you must FACTOR OUT $b$ before reading the phase shift, otherwise you announce $\frac{\pi}{3}$ instead of $\frac{\pi}{6}$ . Amplitude $|a|=3$ ; period $\frac{2\pi}{|b|}=\pi$ ; phase shift $\frac{\pi}{6}$ to the right; vertical shift $1$ . Since $\sin$ runs through $\left[-1, 1\right]$ , the range is $\left[3(-1)+1, \,3(1)+1\right]=\left[-2, 4\right]$ .

d)

$\tan x=\frac{\sin x}{\cos x}$ and $\sec x=\frac{1}{\cos x}$ : both have the same denominator, hence the same domain, $\left\{x\in\mathbb{R}\;:\;x\neq\frac{\pi}{2}+k\pi, \;k\in\mathbb{Z}\right\}$ . What they have in common is precisely the zeros of cosine, which produce a vertical asymptote at each of these values. Remembering this link saves learning two domains: $\tan$ , $\sec$ and their squares all live where $\cos x\neq 0$ , and $\cot$ , $\csc$ where $\sin x\neq 0$ .

Exercise 8: Five statements to correct

Each statement is false. Say in one sentence why, then write the correct statement.

a) “The domain of $\frac{f}{g}$ is the intersection of the domains of $f$ and $g$ .”
b) “The domain of $f\circ g$ is the domain of $g$ .”
c) “ $\sqrt{x^{2}}=x$ .”
d) “Every function has an inverse function: you just have to isolate $x$ .”
e) “The domain of $\sqrt{x^{2}-4}$ is $\left[2, +\infty\right)$ .”

Show the solution

Answers

a) Remove the zeros of $g$
b) $x\in\text{dom}\,g$ and $g(x)\in\text{dom}\,f$
c) $\sqrt{x^{2}}=|x|$
d) Inverse if and only if $f$ is one-to-one
e) $\left(-\infty, -2\right]\cup\left[2, +\infty\right)$

a)

FALSE, a condition is missing. The intersection guarantees that both functions exist, but a quotient also requires that the denominator not be zero. Correct statement: “ $\text{dom}\left(\frac{f}{g}\right)=\left(\text{dom}\,f\cap\text{dom}\,g\right)\setminus\{x:g(x)=0\}$ .” With $f(x)=x$ and $g(x)=x-1$ , the intersection is $\mathbb{R}$ but the domain of the quotient is $\mathbb{R}\setminus\{1\}$ .

b)

FALSE, a condition is missing here too, on the other side. We need $x$ to be in $\text{dom}\,g$ AND $g(x)$ to be in $\text{dom}\,f$ . Correct statement: “ $\text{dom}(f\circ g)=\left\{x\in\text{dom}\,g\;:\;g(x)\in\text{dom}\,f\right\}$ .” With $f(t)=\frac{1}{t-2}$ and $g(x)=\sqrt{x}$ , the domain of $g$ is $\left[0, +\infty\right)$ but that of the composition also excludes $x=4$ .

c)

FALSE for every negative $x$ . $\sqrt{4}=2$ , yet $\sqrt{(-2)^{2}}=\sqrt{4}=2\neq-2$ . Correct statement: “ $\sqrt{x^{2}}=|x|$ .” This is the identity that will produce two horizontal asymptotes in the chapter on limits, and forgetting it there systematically costs half a question.

d)

FALSE. Isolating $x$ is not enough: each $y$ must also have ONLY ONE input, that is, $f$ must be one-to-one. $f(x)=x^{2}$ gives $x=\pm\sqrt{y}$ , two values, so no function. Correct statement: “A function has an inverse if and only if it is one-to-one; otherwise we restrict it to an interval where it is strictly monotonic.” That is what forces the choice of $\left[0, +\infty\right)$ to define the square root, and of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ to define $\arcsin$ .

e)

FALSE, half of the domain is missing. $x^{2}-4\ge 0$ is equivalent to $|x|\ge 2$ , that is, to $x\le-2$ OR $x\ge 2$ . Correct statement: “ $\text{dom}=\left(-\infty, -2\right]\cup\left[2, +\infty\right)$ .” Immediate check: $\sqrt{(-3)^{2}-4}=\sqrt{5}$ does exist. A quadratic with a positive leading coefficient is positive OUTSIDE its roots, and that is the sentence to repeat to yourself before writing an interval.

Exercise 9: Problem: setting up the equation and finding the physical domain

A square of side $x$ is cut from each of the four corners of a rectangular sheet of cardboard measuring $30$ cm by $40$ cm, then the sides are folded up to form an open box. The figure below shows the situation.

a) Express the volume $V$ of the box as a function of $x$ , and give the physically admissible domain.
b) Compute $V(5)$ and $V(12)$ .
c) The formula gives $V(16)=-256$ . What should you conclude about the formula and about the domain?
d) Express the area of cardboard actually used as a function of $x$ , and say on which domain this second function makes sense.

Show the solution

Answers

a) $V(x)=x(30-2x)(40-2x)$ on $\left(0, 15\right)$
b) $3000$ cm³ and $1152$ cm³
c) $x=16$ is outside the domain
d) $S(x)=1200-4x^{2}$ on $\left(0, 15\right)$

a)

After cutting and folding, the height is $x$ , the length $40-2x$ and the width $30-2x$ , because a square is removed at EACH end of each dimension. So $V(x)=x(30-2x)(40-2x)$ . Domain: we need $x>0$ for there to be a height, $30-2x>0$ , that is $x<15$ , and $40-2x>0$ , that is $x<20$ . The most restrictive condition wins: $\text{dom}=\left(0, 15\right)$ . The rule is general: the physical domain is the intersection of ALL the constraints, and it is the smallest one that decides.

b)

$V(5)=5\times 20\times 30=3000$ cm $^{3}$ . $V(12)=12\times 6\times 16=1152$ cm $^{3}$ . The second is smaller even though the cut is larger: the height gained no longer makes up for the base lost, which signals a maximum somewhere between the two and prepares the chapter on optimization.

c)

The formula can still be computed at $x=16$ because a polynomial is defined on all of $\mathbb{R}$ , but it gives $16\times(-2)\times 8=-256$ , a negative volume, which makes no sense. Conclusion: the algebraic expression and the function of the problem are NOT the same object. The function of the problem is the polynomial RESTRICTED to $\left(0, 15\right)$ , and it is the domain, not the formula, that carries the physics. An exam marker takes marks off an optimization answer whose value falls outside the admissible domain, even if the derivative is computed flawlessly.

d)

The area used is the area of the cardboard minus the four squares cut out: $S(x)=1200-4x^{2}$ . It makes sense on the same domain $\left(0, 15\right)$ , to which we may add $x=0$ here, which corresponds to the intact cardboard of area $1200$ cm $^{2}$ but gives no box. Check: $S(15)=1200-900=300$ cm $^{2}$ , that is the two long flaps of $15$ cm by $10$ cm: this is the degenerate case where the width $30-2x$ is zero and the box no longer has a bottom. Comparing the two functions teaches something useful: two quantities from the same problem can have very similar domains without them being identical.

Exercise 10: Problem: a piecewise-defined electricity rate

A utility bills the first $1200$ kilowatt-hours of a billing period at $0.065$ dollars per kilowatt-hour, then each additional kilowatt-hour at $0.10$ dollars. Let $q$ be the consumption for the period, in kilowatt-hours, and $C(q)$ the bill, in dollars.

a) Write $C$ as a piecewise-defined function, and give its domain.
b) Compute $C(900)$ and $C(1500)$ .
c) Show that the graph of $C$ has no jump at the $1200$ kWh threshold, and describe what a rate with a jump would look like.
d) A customer receives a bill of $150$ dollars. What was the consumption? State the question in terms of an inverse function.

Show the solution

Answers

a) $0.065q$ if $q\le 1200$ ; $78+0.10(q-1200)$ otherwise; domain $\left[0, +\infty\right)$
b) $58.50$ dollars and $108$ dollars
c) $78$ on both sides: no jump
d) $C^{-1}(150)=1920$ kWh

a)

Below the threshold, the bill is proportional: $C(q)=0.065q$ for $0\le q\le 1200$ . Beyond it, the customer pays the whole first block, that is $0.065\times 1200=78$ dollars, PLUS the excess at the higher rate: $C(q)=78+0.10(q-1200)$ for $q>1200$ . The domain is $\left[0, +\infty\right)$ . The classic error is to write $0.10q$ in the second piece, which bills the first $1200$ kilowatt-hours at the higher rate and inflates the bill by $42$ dollars: the second piece bills the EXCESS, hence the $q-1200$ .

b)

$900\le 1200$ , so we are in the first piece: $C(900)=0.065\times 900=58.50$ dollars. $1500>1200$ , so in the second: $C(1500)=78+0.10\times 300=78+30=108$ dollars. Choosing the piece BEFORE computing, by comparing $q$ with the threshold, is the whole method.

c)

Compare the two expressions AT the threshold. From the left, $0.065\times 1200=78$ . From the right, $78+0.10(1200-1200)=78$ . Both equal $78$ : the pieces join up, the graph is continuous, and that is exactly what you expect from a fair rate, where using one more kilowatt-hour costs a few cents more and not several dollars at once. A rate with a jump would be, for example, an extra $20$ dollar fee triggered when the threshold is crossed: the bill would then jump suddenly from $78$ to $98$ dollars, and the customer would have every reason to stay below the threshold. Reading this junction is precisely the continuity question of the next chapter, here on an example where the stakes are concrete.

d)

The question asks for the INPUT that $C$ sends to $150$ , that is $C^{-1}(150)$ . Since $150>78$ , we are in the second piece: $78+0.10(q-1200)=150$ gives $0.10(q-1200)=72$ , then $q-1200=720$ and $q=1920$ kWh. Check: $C(1920)=78+0.10\times 720=150$ ✓. The complete inverse is $C^{-1}(y)=\frac{y}{0.065}$ for $0\le y\le 78$ , and $C^{-1}(y)=1200+10(y-78)$ for $y>78$ : a piecewise function has a piecewise inverse, and the thresholds swap, $1200$ on the horizontal axis becoming $78$ on the vertical axis. This inverse exists because $C$ is strictly increasing, which is no detail: a non-monotonic sliding-scale rate would have no inverse.

Exercise 11: The sign chart: when the domain calls for an inequality

Find the domain of each of these functions, in interval notation.

a) $f(x)=\sqrt{\frac{(x+3)(x-1)}{x-2}}$
b) $g(x)=\sqrt[3]{\frac{x-5}{x+1}}$
c) $h(x)=\ln\left(|x|-1\right)$
d) $k(x)=\frac{1}{e^{x}-2}$
e) Compare the domain of $p(x)=\sqrt{x}\cdot\sqrt{x-2}$ with that of $q(x)=\sqrt{x(x-2)}$ .

Show the solution

Answers

a) $\left[-3, 1\right]\cup\left(2, +\infty\right)$
b) $\mathbb{R}\setminus\{-1\}$
c) $\left(-\infty, -1\right)\cup\left(1, +\infty\right)$
d) $\mathbb{R}\setminus\{\ln 2\}$ , $\ln 2\approx 0.693$
e) $\left[2, +\infty\right)$ versus $\left(-\infty, 0\right]\cup\left[2, +\infty\right)$

a)

Two conditions: the radicand is greater than or equal to zero, and the denominator is not zero. The sign of the quotient is read from a chart with three factors, whose zeros $-3$ , $1$ and $2$ split $\mathbb{R}$ into four intervals. For $x<-3$ : $\frac{(-)(-)}{(-)}$ , negative. For $-3<x<1$ : $\frac{(+)(-)}{(-)}$ , positive. For $1<x<2$ : $\frac{(+)(+)}{(-)}$ , negative. For $x>2$ : positive. The quotient is zero at $-3$ and at $1$ , which are kept, and does not exist at $2$ , which is excluded. So $\text{dom}\,f=\left[-3, 1\right]\cup\left(2, +\infty\right)$ . Check: $f(0)=\sqrt{\frac{3\times(-1)}{-2}}=\sqrt{1.5}$ exists, and $f(1.5)$ would require the root of $\frac{4.5\times 0.5}{-0.5}=-4.5$ .

b)

A root with an ODD index accepts a negative radicand: $\sqrt[3]{-8}=-2$ . The only constraint comes from the denominator, $x\neq-1$ . So $\text{dom}\,g=\mathbb{R}\setminus\{-1\}$ . Copying the condition “nonnegative radicand” here would lose the whole interval $\left(-1, 5\right)$ , where $g$ does exist.

c)

The argument of the logarithm must be STRICTLY positive: $|x|-1>0$ , that is $|x|>1$ , so $x<-1$ or $x>1$ . So $\text{dom}\,h=\left(-\infty, -1\right)\cup\left(1, +\infty\right)$ . The endpoints are excluded because $\ln 0$ does not exist.

d)

The only restriction is $e^{x}-2=0$ , that is $e^{x}=2$ and $x=\ln 2\approx 0.693$ . So $\text{dom}\,k=\mathbb{R}\setminus\{\ln 2\}$ . The exponential itself is defined everywhere: it is not what removes a value, the denominator is.

e)

For $p$ , each root requires its own nonnegative radicand: $x\ge 0$ AND $x\ge 2$ , that is $\text{dom}\,p=\left[2, +\infty\right)$ . For $q$ , only the PRODUCT must be nonnegative: $x(x-2)\ge 0$ , that is $x\le 0$ or $x\ge 2$ , and $\text{dom}\,q=\left(-\infty, 0\right]\cup\left[2, +\infty\right)$ . The two expressions agree on $\left[2, +\infty\right)$ but not elsewhere: $q(-1)=\sqrt{3}$ exists while $\sqrt{-1}\cdot\sqrt{-3}$ has no meaning in $\mathbb{R}$ . The identity $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for $a\ge 0$ and $b\ge 0$ , and “simplifying” $p$ into $q$ enlarges the domain without warning.

Exercise 12: Logarithm and exponential: the domain, then the inverse

Consider the functions $u(x)=\ln(2x-6)$ , $v(x)=\ln\left(\frac{x+1}{4-x}\right)$ , $w(x)=\sqrt{\ln x}$ and $p(x)=3+e^{2x-1}$ .

a) Give the domain of $u$ and that of $v$ .
b) Give the domain of $w$ .
c) Give the domain and the range of $p$ .
d) Find $p^{-1}$ and give its domain.
e) Solve $p(x)=5$ exactly, then give the solution to three decimal places.

Show the solution

Answers

a) $\left(3, +\infty\right)$ and $\left(-1, 4\right)$
b) $\left[1, +\infty\right)$
c) Domain $\mathbb{R}$ , range $\left(3, +\infty\right)$
d) $p^{-1}(x)=\frac{1+\ln(x-3)}{2}$ on $\left(3, +\infty\right)$
e) $x=\frac{1+\ln 2}{2}\approx 0.847$

a)

For $u$ : $2x-6>0$ , that is $x>3$ , and $\text{dom}\,u=\left(3, +\infty\right)$ . For $v$ : the quotient must be strictly positive. The numerator changes sign at $-1$ , the denominator at $4$ ; the quotient is positive between them, negative outside, zero at $-1$ and undefined at $4$ . So $\text{dom}\,v=\left(-1, 4\right)$ , open on both sides: at $-1$ we would have $\ln 0$ .

b)

Two layers, two conditions, in order from the inside out. The logarithm requires $x>0$ . The root then requires $\ln x\ge 0$ , that is $x\ge 1$ . The second condition contains the first: $\text{dom}\,w=\left[1, +\infty\right)$ . Stopping at $x>0$ is the typical error: $w(0.5)$ would require $\sqrt{\ln 0.5}=\sqrt{-0.693}$ .

c)

The exponential is defined on $\mathbb{R}$ , so $\text{dom}\,p=\mathbb{R}$ . As $x$ runs through $\mathbb{R}$ , so does $2x-1$ , and $e^{2x-1}$ takes every value in $\left(0, +\infty\right)$ , never $0$ . Adding $3$ : $\text{range}\,p=\left(3, +\infty\right)$ , endpoint excluded. The line $y=3$ is a horizontal asymptote that the curve never reaches.

d)

$p$ is strictly increasing, hence one-to-one. Isolate $x$ in $y=3+e^{2x-1}$ : $e^{2x-1}=y-3$ , possible only if $y>3$ , then $2x-1=\ln(y-3)$ and $x=\frac{1+\ln(y-3)}{2}$ . So $p^{-1}(x)=\frac{1+\ln(x-3)}{2}$ , with domain $\left(3, +\infty\right)$ , which is indeed the range of $p$ : domain and range swap. Check at $x=0.5$ : $p(0.5)=3+e^{0}=4$ and $p^{-1}(4)=\frac{1+\ln 1}{2}=0.5$ .

e)

$3+e^{2x-1}=5$ gives $e^{2x-1}=2$ , so $2x-1=\ln 2$ and $x=\frac{1+\ln 2}{2}\approx 0.847$ . This is exactly $p^{-1}(5)$ : solving $p(x)=5$ and evaluating the inverse at $5$ are one and the same operation. Check: $p(0.847)=3+e^{0.694}\approx 3+2.00=5$ .

Exercise 13: Inverse trigonometric functions: domain, range and the arcsin(sin x) trap

All angle measures are in radians. Recall that $\arcsin$ is the inverse of the restriction of $\sin$ to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , $\arccos$ that of $\cos$ restricted to $\left[0, \pi\right]$ , and $\arctan$ that of $\tan$ restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

a) Give the domain and the range of $\arcsin$ and of $\arccos$ .
b) Compute $\arcsin\left(-\frac{1}{2}\right)$ , $\arccos\left(-\frac{1}{2}\right)$ and $\arctan\sqrt{3}$ .
c) Give the domain of $f(x)=\arcsin(2x-1)$ and that of $g(x)=\arccos\left(1-x^{2}\right)$ .
d) Compute $\arcsin\left(\sin\frac{5\pi}{6}\right)$ and $\sin\left(\arcsin 0.3\right)$ .
e) For which values of $x$ do we have $\arcsin(\sin x)=x$ ?

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

a)

\text{dom}\arcsin

,

Upper endpoint of

\text{range}\arcsin

(in multiples of π)

Upper endpoint of

\text{range}\arccos

(in multiples of π)

b)

\arcsin\left(-\frac{1}{2}\right)

(in multiples of π)

\arccos\left(-\frac{1}{2}\right)

(in multiples of π)

\arctan\sqrt{3}

(in multiples of π)

c)

\text{dom}\,f

,

Upper endpoint of

\text{dom}\,g

d)

\arcsin\left(\sin\frac{5\pi}{6}\right)

(in multiples of π)

\sin\left(\arcsin 0.3\right)

e)

Left endpoint (in multiples of π)

Right endpoint (in multiples of π)

Show the solution

Answers

a) $\arcsin$ : $\left[-1, 1\right]\to\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ; $\arccos$ : $\left[-1, 1\right]\to\left[0, \pi\right]$
b) $-\frac{\pi}{6}$ ; $\frac{2\pi}{3}$ ; $\frac{\pi}{3}$
c) $\left[0, 1\right]$ and $\left[-\sqrt{2}, \sqrt{2}\right]$
d) $\frac{\pi}{6}$ and $0.3$
e) $x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

a)

The inverse swaps the domain and the range of the RESTRICTION. $\sin$ restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ takes its values in $\left[-1, 1\right]$ : $\text{dom}\arcsin=\left[-1, 1\right]$ and $\text{range}\arcsin=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Likewise, $\text{dom}\arccos=\left[-1, 1\right]$ and $\text{range}\arccos=\left[0, \pi\right]$ . The two ranges differ because the restrictions differ: that is the source of half the errors of the chapter.

b)

$\arcsin\left(-\frac{1}{2}\right)$ is the unique angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $-\frac{1}{2}$ : $-\frac{\pi}{6}$ . $\arccos\left(-\frac{1}{2}\right)$ is the unique angle in $\left[0, \pi\right]$ whose cosine is $-\frac{1}{2}$ : $\frac{2\pi}{3}$ , and not $-\frac{\pi}{3}$ or $\frac{4\pi}{3}$ , which lie outside the range. $\arctan\sqrt{3}=\frac{\pi}{3}$ .

c)

The argument of $\arcsin$ must be in $\left[-1, 1\right]$ : $-1\le 2x-1\le 1$ gives $0\le x\le 1$ , that is $\text{dom}\,f=\left[0, 1\right]$ . For $g$ : $-1\le 1-x^{2}\le 1$ . The right-hand inequality, $x^{2}\ge 0$ , is always true; the left-hand one gives $x^{2}\le 2$ , that is $-\sqrt{2}\le x\le\sqrt{2}$ . So $\text{dom}\,g=\left[-\sqrt{2}, \sqrt{2}\right]$ , with $\sqrt{2}\approx 1.414$ .

d)

$\sin\frac{5\pi}{6}=\frac{1}{2}$ , and $\arcsin\frac{1}{2}=\frac{\pi}{6}$ . So $\arcsin\left(\sin\frac{5\pi}{6}\right)=\frac{\pi}{6}$ , and not $\frac{5\pi}{6}$ : $\arcsin$ always returns an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and $\frac{5\pi}{6}$ is not in it. In the other order, $0.3$ is in the domain of $\arcsin$ , and $\sin\left(\arcsin 0.3\right)=0.3$ exactly.

e)

$\arcsin(\sin x)=x$ first requires $x$ to be a possible value of $\arcsin$ , so $x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , and on this interval the identity holds since $\sin$ there is the function that was inverted. Conclusion: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . On the other hand, $\sin(\arcsin x)=x$ holds for every $x$ in $\left[-1, 1\right]$ . The rule: $f\left(f^{-1}(x)\right)=x$ on the domain of $f^{-1}$ , but $f^{-1}\left(f(x)\right)=x$ only on the chosen restriction.

Exercise 14: Problem: the spreading slick, composing two models

A leak forms a circular slick on the surface of a lake; its radius, in metres, is $r(t)=2+3t$ , where $t$ is the time in hours since the first measurement. The leak is sealed after $8$ hours and the model then stops applying. The area of a disk of radius $r$ is $A(r)=\pi r^{2}$ . The figure below shows the situation.

a) Which of the two compositions, $A\circ r$ or $r\circ A$ , gives the area of the slick at time $t$ ? Write it and give its domain in context.
b) Compute the area of the slick after $2$ hours.
c) Give the range of $A\circ r$ on its domain.
d) At what moment does the area reach $400$ m $^{2}$ ?
e) Justify that $A\circ r$ has an inverse on its domain, write it and give its domain.

Show the solution

Answers

a) $(A\circ r)(t)=\pi(2+3t)^{2}$ on $\left[0, 8\right]$
b) $64\pi\approx 201$ m²
c) $\left[4\pi, 676\pi\right]\approx\left[12.57, 2124\right]$
d) $t\approx 3.09$ h
e) $t=\frac{1}{3}\left(\sqrt{\frac{S}{\pi}}-2\right)$ on $\left[4\pi, 676\pi\right]$

a)

We first know the radius at time $t$ , and from it we get the area: $r$ acts first, $A$ second, and that is $A\circ r$ . $(A\circ r)(t)=\pi(2+3t)^{2}$ . The other composition, $r\left(A(x)\right)=2+3\pi x^{2}$ , makes no sense here: it would treat an area as a time. The domain comes from the context, not from the formula: $t\in\left[0, 8\right]$ , whereas the polynomial is defined on $\mathbb{R}$ .

b)

$r(2)=8$ m, so $A(8)=64\pi\approx 201$ m $^{2}$ .

c)

On $\left[0, 8\right]$ , $r$ increases from $r(0)=2$ to $r(8)=26$ , and $A$ is increasing for $r\ge 0$ : the area increases from $4\pi\approx 12.57$ m $^{2}$ to $676\pi\approx 2124$ m $^{2}$ . The range is $\left[4\pi, 676\pi\right]$ . Careful, the area does not start at $0$ : at $t=0$ , the slick already has a $2$ m radius.

d)

$\pi(2+3t)^{2}=400$ gives $(2+3t)^{2}=\frac{400}{\pi}$ , and since $2+3t>0$ , $2+3t=\sqrt{\frac{400}{\pi}}\approx 11.284$ . So $t=\frac{11.284-2}{3}\approx 3.09$ hours, about $3$ h $6$ min. Only the positive root is kept: the negative root would give a negative radius, and a negative time outside the domain.

e)

On $\left[0, 8\right]$ , $A\circ r$ is strictly increasing, as a composition of two functions that are increasing on the sets involved: it is one-to-one, hence invertible. Isolating $t$ in $S=\pi(2+3t)^{2}$ : $t=\frac{1}{3}\left(\sqrt{\frac{S}{\pi}}-2\right)$ . Its domain is the range found in c), $\left[4\pi, 676\pi\right]\approx\left[12.57, 2124\right]$ . This inverse answers the practical question “how long since the first measurement, given the surface area seen from a plane?”, and part d) was a special case of it.

Exercise 15: Problem: the arched tunnel, transformations and clearance

The vault of a tunnel has the profile of the parabola $y=6-0.24x^{2}$ , in metres, where $x$ is the horizontal distance from the axis of the tunnel and the ground is at $y=0$ . The figure shows this profile and a truck $3$ m wide centred in the tunnel.

a) Give the width of the tunnel at ground level, then the domain and the range of the function that describes the vault.
b) A truck $4$ m high goes through the tunnel. What width of roadway is left at that height?
c) The roadway is resurfaced and rises by $0.3$ m. Write the height of the vault above the new roadway, its domain, and the width available at a height of $4$ m.
d) To widen the tunnel, the original vault is replaced by its horizontal stretch by a factor of $1.2$ , with the height at the centre unchanged. Write the new profile and its width at ground level.
e) In the original tunnel, what is the maximum height of a truck $3$ m wide driving in the centre?

Show the solution

Answers

a) $10$ m; domain $\left[-5, 5\right]$ , range $\left[0, 6\right]$
b) $5.77$ m
c) $h(x)=5.7-0.24x^{2}$ on $\left[-4.87, 4.87\right]$ ; $5.32$ m
d) $y=6-\frac{x^{2}}{6}$ ; $12$ m at ground level
e) $5.46$ m

a)

The vault meets the ground when $6-0.24x^{2}=0$ , that is $x^{2}=25$ and $x=\pm 5$ : the tunnel is $10$ m wide at ground level. The function of the vault only has physical meaning between its two feet: domain $\left[-5, 5\right]$ . On this domain, the height goes from $0$ at the feet to $6$ at the centre: range $\left[0, 6\right]$ .

b)

Find where the vault is at $4$ m: $6-0.24x^{2}=4$ gives $x^{2}=\frac{2}{0.24}\approx 8.333$ and $x\approx\pm 2.887$ . The width available at that height is $2\times 2.887\approx 5.77$ m. Answering $2.89$ m, the half-width, is the most frequent error: $x$ is measured from the axis.

c)

Everything happens as if the vault came down by $0.3$ m: it is a vertical translation, $h(x)=5.7-0.24x^{2}$ . New domain: $0.24x^{2}\le 5.7$ , that is $x^{2}\le 23.75$ and $|x|\le 4.873$ , so $\left[-4.87, 4.87\right]$ to two decimal places. A VERTICAL translation has changed the domain, because the domain is defined here by the condition $h\ge 0$ . At $4$ m: $0.24x^{2}=1.7$ , $x^{2}\approx 7.083$ , $|x|\approx 2.661$ , that is a width of $5.32$ m. Thirty centimetres of asphalt cost almost half a metre of usable width.

d)

A horizontal stretch by a factor of $1.2$ replaces $x$ with $\frac{x}{1.2}$ : $y=6-0.24\left(\frac{x}{1.2}\right)^{2}=6-\frac{0.24}{1.44}x^{2}=6-\frac{x^{2}}{6}$ . At ground level, $x^{2}=36$ and $x=\pm 6$ : the width becomes $12$ m, that is $10\times 1.2$ . Writing $6-0.24(1.2x)^{2}$ would COMPRESS the tunnel: what touches $x$ acts the opposite way.

e)

The truck occupies $x\in\left[-1.5, 1.5\right]$ . Since the vault is lower at the sides, the constraint is read at the edge of the truck: $6-0.24\times 1.5^{2}=6-0.54=5.46$ m. A truck $3$ m wide must be less than $5.46$ m high, and not $6$ m, a height that only holds at the centre, over a width of zero.

Corrected exercises: functions, domains and operations (201-SN2-RE)

Course recap

Calculations: domains to find

Exercise 0.1: Find the domain: twelve direct calculations

Calculations: compositions and inverses to evaluate

Exercise 0.2: Compose two functions and invert: ten direct calculations

Part A: domain, operations and composition (/50)

Exercise 1: The domain: the only three restrictions

Exercise 2: Absolute value and piecewise-defined function

Exercise 3: Operations on functions: the domain does not simply carry over

Exercise 4: Composition, and the domain that is not the one you expect

Exercise 5: Reading a transformed graph

Part B: inverses, models and reasoning (/50)

Exercise 6: Inverse functions: existence, domain and range

Exercise 7: The unit circle and periodic functions

Exercise 8: Five statements to correct

Exercise 9: Problem: setting up the equation and finding the physical domain

Exercise 10: Problem: a piecewise-defined electricity rate

Part C: the classics (/50)

Exercise 11: The sign chart: when the domain calls for an inequality

Exercise 12: Logarithm and exponential: the domain, then the inverse

Exercise 13: Inverse trigonometric functions: domain, range and the arcsin(sin x) trap

Exercise 14: Problem: the spreading slick, composing two models

Exercise 15: Problem: the arched tunnel, transformations and clearance

See also

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