MATH 203 functions, trig and exponentials: revision sheet

The thread of the chapter

Rewrite first, read after: factor the inside before reading a shift ( $3x - 6 = 3(x - 2)$ ), turn every root into a rational exponent and every negative exponent into a reciprocal before simplifying, bring an angle back to its reference angle and its quadrant before reading a value, and FACTOR a trigonometric equation instead of dividing it; every mark of this chapter is lost on the unrewritten form.

Topics covered

Composition of functions
Horizontal and vertical shifts
Horizontal and vertical scaling
Radian measure
Reference angle
Six trigonometric functions
Pythagorean identity
Double-angle formulas
Trigonometric equations
General sine function
Laws of exponents
Exponential functions

The essentials

A trigonometric value: size from the reference angle, sign from the quadrant

• Radians first: $\pi$ rad $= 180^\circ$ , and $s = r\theta$ holds only with $\theta$ in radians. In calculus a number with no unit is an angle in radians.
• The three angles to know: $\sin\frac{\pi}{6} = \frac{1}{2}$ , $\sin\frac{\pi}{4} = \frac{\sqrt 2}{2}$ , $\sin\frac{\pi}{3} = \frac{\sqrt 3}{2}$ , and the cosines in the reverse order.
• Reference angle: the acute angle between the terminal side and the $x$ -axis. $\frac{5\pi}{6}$ , $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ all have reference angle $\frac{\pi}{6}$ .
• Signs: $\sin > 0$ in quadrants I and II, $\cos > 0$ in I and IV, $\tan > 0$ in I and III. A right triangle gives SIZES only.
• Remove full turns first: $2\pi$ for $\sin$ , $\cos$ , $\sec$ , $\csc$ ; $\pi$ for $\tan$ and $\cot$ . $\cos$ is even, $\sin$ and $\tan$ are odd.

Four angles, ONE reference angle

\frac{\pi}{6}

: the four points have the same coordinates up to sign, and the quadrant alone decides the signs.

Write the reference angle and the quadrant on your copy before the value: “reference $\frac{\pi}{6}$ , quadrant III, so $\sin\frac{7\pi}{6} = -\frac{1}{2}$ .” The line earns the method mark even if the arithmetic slips.

The rewrites the whole chapter runs on

• Inside a transformation, factor the coefficient of $x$ : $f(3x - 6) = f(3(x - 2))$ is a compression by $3$ and a shift of $2$ , not $6$ .
• Outside, the order of operations of the formula: in $-2f(x) + 1$ , multiply by $-2$ first, then add $1$ .
• Roots become rational exponents, $\sqrt[n]{x^m} = x^{m/n}$ ; negative exponents become reciprocals, $x^{-r} = \frac{1}{x^r}$ , never negative numbers.
• $\sin 2x = 2\sin x\cos x$ ; $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ ; $\sin^2 x = \frac{1 - \cos 2x}{2}$ ; $\sin^2 x + \cos^2 x = 1$ .
• A composition $f(g(x))$ : domain of $g$ first, then $g(x)$ in the domain of $f$ , both decided BEFORE any simplification.

y = -2\sin x + 1

: multiply by

-2

first (reflect, stretch), then add

1

. The midline moves to

y = 1

and the values run from

-1

3

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Rewrites that exist, and rewrites that do not

Read a line as: the expression of the first column, by the law of the second, becomes the form of the third, checked at a number underneath. The red lines are rewrites that students write and that are false.

Expression	Law	Rewritten
$\sqrt[4]{x^6}$	$\sqrt[n]{a^m} = a^{m/n}$	$x^{3/2}$
Example: At $x = 4$ : $\sqrt[4]{4096} = 8$ and $4^{3/2} = 8$ .
$\frac{1}{\sqrt[3]{x^2}}$	$\frac{1}{a^r} = a^{-r}$	$x^{-2/3}$
Example: At $x = 8$ : $\frac{1}{4}$ both ways.
$\left(x^{-2/3}\right)^3$	$(a^r)^s = a^{rs}$	$x^{-2}$
Example: At $x = 8$ : $\left(\frac{1}{4}\right)^3 = \frac{1}{64} = 8^{-2}$ .
$2^{x + 3}$	$a^{x + y} = a^x a^y$	$8 \cdot 2^x$
Example: At $x = 1$ : $2^4 = 16 = 8 \times 2$ .
$\sin 2x$	double angle	$2\sin x\cos x$
Example: At $x = \frac{\pi}{6}$ : $\sin\frac{\pi}{3} = \frac{\sqrt 3}{2} = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt 3}{2}$ .
$\sin 2x$	“double it”	$2\sin x$ rule that does not exist
Example: At $x = \frac{\pi}{2}$ : $\sin\pi = 0$ , while $2\sin\frac{\pi}{2} = 2$ . What to do: Expand with $\sin 2x = 2\sin x\cos x$ ; a sine is never larger than $1$ .
$x^{-2/3}$	“minus sign out”	$-\sqrt[3]{x^2}$ rule that does not exist
Example: At $x = 8$ : $8^{-2/3} = \frac{1}{4}$ , while $-\sqrt[3]{64} = -4$ . What to do: A negative exponent is a RECIPROCAL: $x^{-2/3} = \frac{1}{\sqrt[3]{x^2}}$ .
$(x + 16)^{1/2}$	“term by term”	$x^{1/2} + 4$ rule that does not exist
Example: At $x = 9$ : $\sqrt{25} = 5$ , while $3 + 4 = 7$ . What to do: Exponent laws act on PRODUCTS; a power of a sum stays a power of a sum.

Every line is checked the same way: pick a number where both sides are easy, $x = 4$ , $x = 8$ , $x = \frac{\pi}{2}$ , and compare.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading the shift before factoring the inside

2 marks: every point of the sketch is misplaced

What not to write

“ $y = f(3x - 6)$ : compress by $3$ , then shift $6$ units right.”

What to write

“ $f(3x - 6) = f(3(x - 2))$ : compress by $3$ , then shift $2$ units right. The point $(0, 2)$ of $f$ goes to $x = \frac{0 + 6}{3} = 2$ .”

y = \sqrt{3x - 6} = \sqrt{3(x - 2)}

starts at

x = 2

(green). Compress by

3

then shift

6

gives

\sqrt{3(x - 6)}

, which starts at

6

(red): another function.

Why: After the compression the shift acts on the compressed graph, so it is read on $x - 2$ . Compressing and THEN shifting by $6$ graphs $f(3x - 18)$ .

2. An arc length computed in degrees

the whole question

What not to write

“The arc cut by $150^\circ$ on a circle of radius $6$ cm is $s = 6 \times 150 = 900$ cm.”

What to write

“ $150^\circ = \frac{5\pi}{6}$ rad, so $s = r\theta = 6 \times \frac{5\pi}{6} = 5\pi \approx 15.71$ cm.”

The arc cut by

150^\circ

on a circle of radius

6

cm is

s = r\theta = 6 \cdot \frac{5\pi}{6} = 5\pi \approx 15.71

cm, less than the circumference

12\pi \approx 37.7

cm;

900

cm is impossible.

Why: $s = r\theta$ is the DEFINITION of the radian, so it holds in radians only. $900$ cm is more than twenty times the whole circumference, $12\pi \approx 37.7$ cm.

3. Taking the signs from the triangle

1 mark per value, often the whole part

What not to write

“ $\tan\theta = \frac{15}{8}$ with $\pi < \theta < \frac{3\pi}{2}$ : the triangle $8$ , $15$ , $17$ gives $\sin\theta = \frac{15}{17}$ and $\cos\theta = \frac{8}{17}$ .”

What to write

“The triangle gives the sizes $\frac{15}{17}$ and $\frac{8}{17}$ ; in quadrant III sine and cosine are negative: $\sin\theta = -\frac{15}{17}$ , $\cos\theta = -\frac{8}{17}$ .”

The triangle gives the SIZES

8

15

17

; the point

(-8, -15)

of quadrant III gives the signs:

\sin\theta = -\frac{15}{17}

\cos\theta = -\frac{8}{17}

Why: A triangle has positive sides; the signs live on the unit circle. The given tangent is positive in quadrants I and III alike, so only the interval can decide.

4. Leaving the calculator in degree mode

every numerical answer that follows

What not to write

“ $\cos\frac{\pi}{3} = 0.9998$ (calculator).”

What to write

“ $\cos\frac{\pi}{3} = \frac{1}{2}$ exactly; in radian mode the calculator agrees.”

Why: In degree mode the calculator takes the cosine of $1.047$ DEGREES, a tiny angle, hence a value near $1$ . Checking one known value, $\cos\frac{\pi}{3} = 0.5$ , before an exam question costs five seconds.

5. Dividing a trigonometric equation by cos x

half the question: two solutions out of four

What not to write

“ $\sin 2x = \cos x$ , so $2\sin x\cos x = \cos x$ , so $2\sin x = 1$ : $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .”

What to write

“ $2\sin x\cos x - \cos x = \cos x\,(2\sin x - 1) = 0$ : $\cos x = 0$ or $\sin x = \frac{1}{2}$ , so $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ .”

The curves

y = \sin 2x

and

y = \cos x

meet FOUR times on

[0, 2\pi]

: the two green points survive the division, the two red points on the axis do not.

Why: Dividing by $\cos x$ is legal only where $\cos x \ne 0$ , and the solutions it deletes are exactly those where $\cos x = 0$ . Factoring keeps them.

6. Forgetting that 2x runs over a doubled interval

2 marks: half the solutions

What not to write

“ $2\cos 2x = 1$ on $[0, 2\pi]$ : $2x = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ , so $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .”

What to write

“With $u = 2x$ , $u$ runs over $[0, 4\pi]$ : $u = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$ , so $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ .”

[0, 2\pi]

y = 2\cos 2x

makes TWO full cycles and meets

y = 1

four times:

\frac{\pi}{6}

and

\frac{5\pi}{6}

(green), and also

\frac{7\pi}{6}

and

\frac{11\pi}{6}

(red), lost when

2x

stays in

[0, 2\pi]

Why: When $x$ makes one turn, $2x$ makes two. Write the interval of the new angle BEFORE going to the unit circle.

7. Taking half a period for a period

2 marks:

B

C

and the sketch are all wrong

What not to write

“The sinusoid has a maximum at $x = 1$ and a minimum at $x = 4$ , so its period is $3$ and $B = \frac{2\pi}{3}$ .”

What to write

“From a maximum to the next minimum is HALF a period: the period is $2 \times 3 = 6$ and $B = \frac{2\pi}{6} = \frac{\pi}{3}$ .”

From the maximum at

1

to the minimum at

4

is HALF a cycle (red); the next maximum is at

7

, so the period is

6

(green) and

B = \frac{\pi}{3}

Why: A full cycle goes maximum, midline, minimum, midline, maximum. The quarter period, from the midline to the maximum, is what places $C$ .

8. Reading a domain on the simplified formula

1 mark, and every later answer that uses the domain

What not to write

“ $f(x) = \sqrt{x - 1}$ , so $(f \cdot f)(x) = x - 1$ , whose domain is $\mathbb{R}$ .”

What to write

“ $(f \cdot f)(x) = x - 1$ for $x \ge 1$ only: the domain of $f \cdot f$ is $D(f) \cap D(f) = [1, \infty)$ , since $f(0)$ does not exist.”

(f \cdot f)(x) = x - 1

holds only where

f(x) = \sqrt{x - 1}

exists: the graph is the green half-line from

(1, 0)

, not the whole line.

Why: The domain belongs to the construction, not to the final formula. Simplifying $\sqrt{x - 1}\,\sqrt{x - 1}$ assumed $x \ge 1$ ; the result forgets it, the function does not.

Which method to choose

Which rewrite solves a trigonometric equation, by its FORM

Look at which functions and which angles appear, then rewrite until one function of one angle is left

If one function of one angle equals a number, $\cos 2x = \frac{1}{2}$ → isolate, name $u$ the angle, compute the interval of $u$ , then read the unit circle

Example: $x \in [0, 2\pi]$ gives $u = 2x \in [0, 4\pi]$ : four solutions
If $\sin 2x$ with $\sin x$ or $\cos x$ → expand $\sin 2x = 2\sin x\cos x$ , bring everything to one side, FACTOR the common factor

Example: $\sin 2x = \cos x$ : $\cos x\,(2\sin x - 1) = 0$
If $\sin^2 x$ together with $\cos x$ → replace $\sin^2 x$ by $1 - \cos^2 x$ : a quadratic in $\cos x$

Example: $2\sin^2 x + 3\cos x = 3$ : $(2\cos x - 1)(\cos x - 1) = 0$
If $\cos 2x$ together with $\sin x$ → use $\cos 2x = 1 - 2\sin^2 x$ : a quadratic in $\sin x$

Example: $\cos 2x = \sin x$ : $\sin x = \frac{1}{2}$ or $-1$ , so $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$
If fourth powers, $\cos^4 x - \sin^4 x$ → difference of squares, then $\sin^2 x + \cos^2 x = 1$

Example: $\cos^4 x - \sin^4 x = \cos 2x$

No branch ever divides by $\sin x$ or $\cos x$ . If none applies, write everything in $\sin x$ and $\cos x$ and look for a common factor or a Pythagorean pair; the inverse trigonometric keys belong to chapter 13 and are never needed for the exact values of this chapter.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a trigonometric equation on an interval

When to use it: “Solve on $[0, 2\pi]$ ”, “find all $x$ in the interval such that”

1 Rewrite with one function of one angle, using the identity named by the form (double angle, Pythagoras), and bring everything to one side.
2 Factor, and set each factor equal to $0$ ; never divide by an expression that can vanish.
3 For each factor, name the angle $u$ and compute the interval that $u$ runs over when $x$ runs over the given interval.
4 Read every solution of $u$ on the unit circle within that interval, including the endpoints if the interval is closed, then go back to $x$ .
5 Check one solution in the ORIGINAL equation, and state the list.

Concluding sentence

“ $\cos x\,(2\sin x - 1) = 0$ , so $\cos x = 0$ or $\sin x = \frac{1}{2}$ . On $[0, 2\pi]$ , the solutions are $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ .”

The trap: Dividing by $\cos x$ , and solving $\cos 2x = c$ on one turn when $2x$ makes two.

Marking: Typically 1 mark for the rewrite, 1 for the factorization, and 1 per solution found; a missing solution costs its mark, a spurious one too.

Writing a sinusoid from its graph

When to use it: A graph with a marked maximum and minimum, and “find $A$ , $B$ , $C$ , $D$ ”

1 Midline: $D = \frac{\max + \min}{2}$ . Amplitude: $A = \frac{\max - \min}{2}$ .
2 Period: twice the horizontal distance from a maximum to the next minimum; then $B = \frac{2\pi}{\text{period}}$ .
3 Choose the function: for $\sin$ with $A > 0$ , $C$ is where the curve crosses the midline going UP, a quarter period before a maximum; for $\cos$ , $C$ is at a maximum.
4 Check the formula at the marked maximum.

Concluding sentence

“ $D = \frac{5 + (-1)}{2} = 2$ , $A = 3$ , the half period is $4 - 1 = 3$ , so the period is $6$ and $B = \frac{\pi}{3}$ ; the maximum is at $x = 1$ , so $g(x) = 3\cos\left(\frac{\pi}{3}(x - 1)\right) + 2$ .”

The trap: Taking the distance from a maximum to a minimum as the whole period.

Marking: Typically 1 mark each for A, B, C and D, and 1 for a formula that passes the check.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Each solution in the original equation

Substitute every solution you found into the equation as it was GIVEN, not into a rewritten form.

$x = \frac{\pi}{2}$ in $\sin 2x = \cos x$ : $\sin\pi = 0 = \cos\frac{\pi}{2}$ , so it IS a solution, even though the divided equation $2\sin x = 1$ rejects it.
The mode, on one known value

Before the first trigonometric computation, type the cosine of pi over 3.

$0.5$ means radian mode; $0.9998$ means degree mode, and every value computed so far must be redone.
One point through a transformed formula

Take a named point of the original graph, compute its image, and plug the image's x into the new formula.

For $y = -2f(3x - 6) + 1$ and $B(0, 2)$ : image $(2, -3)$ , and $-2f(3 \cdot 2 - 6) + 1 = -2f(0) + 1 = -3$ . An image at $x = 6$ would give $f(12)$ , outside the domain: the error shows.
An exponent rewrite at x = 8

Evaluate the original and the rewritten power at a number whose roots are easy, 4, 8 or 64.

$\frac{1}{\sqrt[3]{x^2}}$ at $x = 8$ is $\frac{1}{4}$ ; if your rewrite gives $-4$ , the negative exponent became a negative sign.

The typical problem, taken apart

A sinusoid read from its graph, then an equation on an interval

The figure shows the graph of a sinusoidal function $f$ on $[0, \pi]$ , with a maximum at $\left(\frac{5\pi}{12}, 1\right)$ and a minimum at $\left(\frac{11\pi}{12}, -3\right)$ . Write $f(x) = A\sin(B(x - C)) + D$ with $A > 0$ , $B > 0$ , then solve $f(x) = 0$ on $[0, \pi]$ .

Exact answers, as on a MATH 203 midterm.

The maximum and the minimum are the only data needed: the midline, the amplitude and the half period are read between them.

Step 1

$D = \frac{1 + (-3)}{2} = -1$ and $A = \frac{1 - (-3)}{2} = 2$ .

Why

The midline is halfway between the extremes, the amplitude is the distance from the midline to an extreme. Both come before anything horizontal.

Step 2

Half period: $\frac{11\pi}{12} - \frac{5\pi}{12} = \frac{\pi}{2}$ , so the period is $\pi$ and $B = \frac{2\pi}{\pi} = 2$ .

Why

From a maximum to the next minimum is HALF a cycle; doubling it is the step that is most often skipped.

Step 3

The sine starts its cycle on the midline going up, a quarter period ( $\frac{\pi}{4}$ ) before the maximum: $C = \frac{5\pi}{12} - \frac{3\pi}{12} = \frac{\pi}{6}$ . So $f(x) = 2\sin\left(2\left(x - \frac{\pi}{6}\right)\right) - 1$ .

Why

$C$ is where the sine STARTS, not where the maximum is; with a cosine it would be $\frac{5\pi}{12}$ directly.

Step 4

$f(x) = 0$ gives $\sin u = \frac{1}{2}$ with $u = 2\left(x - \frac{\pi}{6}\right)$ . For $x \in [0, \pi]$ , $u \in \left[-\frac{\pi}{3}, \frac{5\pi}{3}\right]$ , where $\sin u = \frac{1}{2}$ at $u = \frac{\pi}{6}$ and $u = \frac{5\pi}{6}$ only.

Why

The interval of $u$ is computed from the interval of $x$ ; the next solution, $\frac{13\pi}{6}$ , is beyond $\frac{5\pi}{3}$ and is correctly excluded.

Step 5

Back to $x = \frac{u}{2} + \frac{\pi}{6}$ : $x = \frac{\pi}{12} + \frac{2\pi}{12} = \frac{\pi}{4}$ and $x = \frac{5\pi}{12} + \frac{2\pi}{12} = \frac{7\pi}{12}$ . Check: $f\left(\frac{\pi}{4}\right) = 2\sin\frac{\pi}{6} - 1 = 0$ .

Why

Undoing the inside is done in the reverse order, divide by $2$ , then add $\frac{\pi}{6}$ ; the check in $f$ confirms both the formula and the solution.

The conclusion, written out

“ $f(x) = 2\sin\left(2\left(x - \frac{\pi}{6}\right)\right) - 1$ , and on $[0, \pi]$ the solutions of $f(x) = 0$ are $x = \frac{\pi}{4}$ and $x = \frac{7\pi}{12}$ .”

The classic mistake on this problem: Taking the period as $\frac{\pi}{2}$ , which gives $B = 4$ , or placing $C$ at the maximum $\frac{5\pi}{12}$ with a sine, which shifts the whole graph by a quarter period.

Learn by heart

• $\pi$ rad $= 180^\circ$ ; $s = r\theta$ in RADIANS only. The calculator stays in radian mode.
• Trigonometric value: size from the reference angle, sign from the quadrant. A triangle gives sizes, never signs.
• $\sin 2x = 2\sin x\cos x$ , $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ , $\sin^2 x + \cos^2 x = 1$ .
• Trigonometric equation: one function of one angle, FACTOR, never divide; the angle $2x$ runs over a doubled interval.
• $y = A\sin(B(x - C)) + D$ : period $\frac{2\pi}{|B|}$ ; maximum to minimum is HALF a period.
• $f(bx - d) = f\left(b\left(x - \frac{d}{b}\right)\right)$ : compress by $b$ , shift by $\frac{d}{b}$ .
• $x^{-r} = \frac{1}{x^r}$ , $\sqrt[n]{x^m} = x^{m/n}$ ; exponent laws act on products, never on sums.

Frequently asked questions

Why does my calculator give the wrong value for cos(pi/3) in MATH 203?

It is almost always in degree mode. It then takes the cosine of about 1.047 degrees, a tiny angle, and returns a value close to 1, such as 0.9998, instead of one half. In calculus every angle is measured in radians, so switch the calculator to radian mode at the start of the exam and check it on the cosine of pi over 3, which must give 0.5.

How do I find the horizontal shift of y = sin(2x - pi/2)?

Factor the coefficient of x out of the angle first: 2x minus pi over 2 equals 2 times the quantity x minus pi over 4. The graph is the sine compressed horizontally by 2 and shifted pi over 4 to the right, not pi over 2. The shift can only be read once the inside has the form of a number times x minus the shift.

Why can I not divide both sides of a trigonometric equation by cos x?

Because cos x can be zero, and the values where it is zero may be solutions. Dividing deletes them without warning. In sin 2x equals cos x, dividing leaves only two of the four solutions on one turn. Bring everything to one side, factor out the common cos x, and set each factor equal to zero instead.

Is x to the power minus 2/3 a negative number?

No. A negative exponent means a reciprocal: x to the minus two thirds is one over the cube root of x squared. For x equal to 8 it is one quarter, a positive number. For positive x every power of x is positive, whatever the sign of the exponent, so a negative answer is the sign of a rewriting error.

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Corrected exercises: Functions, trigonometric and exponential functions, MATH 203 at Concordia

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150 minutes

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Revision sheet: functions, trigonometric and exponential functions (MATH 203)

The essentials

A trigonometric value: size from the reference angle, sign from the quadrant

The rewrites the whole chapter runs on

The rules in table form

Rewrites that exist, and rewrites that do not

The mistakes that cost marks

1. Reading the shift before factoring the inside

2. An arc length computed in degrees

3. Taking the signs from the triangle

4. Leaving the calculator in degree mode

5. Dividing a trigonometric equation by cos x

6. Forgetting that 2x runs over a doubled interval

7. Taking half a period for a period

8. Reading a domain on the simplified formula

Which method to choose

Which rewrite solves a trigonometric equation, by its FORM

How the answer is expected to be written

Solving a trigonometric equation on an interval

Writing a sinusoid from its graph

Check before you hand in

The typical problem, taken apart

A sinusoid read from its graph, then an equation on an interval

Learn by heart

Frequently asked questions

Why does my calculator give the wrong value for cos(pi/3) in MATH 203?

How do I find the horizontal shift of y = sin(2x - pi/2)?

Why can I not divide both sides of a trigonometric equation by cos x?

Is x to the power minus 2/3 a negative number?

Corrected exercises: Functions, trigonometric and exponential functions, MATH 203 at Concordia

See also

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