Functions, graphs and inverses

Everything later in the course is applied to functions, so the term opens by making sure you can move between a formula, a graph and a description in words. You work with domain and range, compose and invert functions, predict what a shift, stretch or reflection does to a graph, and solve the exponential and trigonometric equations that later chapters take for granted.

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In this topic

Taught: Before the midterm
Questions: 19
Total marks: 45
Suggested time: 74 min
Difficulty: Introductory to Capstone
Show the 3 hardest

Key results

Composition: $(f \circ g)(x) = f\bigl(g(x)\bigr)$
For $x$ in the domain of $g$ with $g(x)$ in the domain of $f$ .
The inverse function: $f^{-1}(y) = x \iff f(x) = y$
When $f$ is one-to-one; the domain of $f^{-1}$ is then the range of $f$ , and its range is the domain of $f$ .
Shifts and reflections: $\begin{aligned} y &= f(x - c) + d && \text{moved } c \text{ right and } d \text{ up} \\ y &= -f(x) && \text{reflected in the } x\text{-axis} \\ y &= f(-x) && \text{reflected in the } y\text{-axis} \end{aligned}$
For constants $c$ and $d$ , each compared with the graph of $y = f(x)$ ; a negative $c$ or $d$ moves it left or down.
Laws of logarithms: $\log(ab) = \log a + \log b, \qquad \log\frac{a}{b} = \log a - \log b, \qquad \log\bigl(a^{r}\bigr) = r\log a$
For $a > 0$ , $b > 0$ and any constant $r$ .
Exponential and logarithm: $e^{\log x} = x, \qquad \log\bigl(e^{y}\bigr) = y$
Here $\log$ is the natural logarithm, base $e$ : the first holds for every $x > 0$ , the second for every $y$ .
Trigonometric identities: $\tan x = \frac{\sin x}{\cos x}, \qquad \sec x = \frac{1}{\cos x}, \qquad \sin^{2}x + \cos^{2}x = 1$
The identity holds for every $x$ ; $\tan x$ and $\sec x$ are defined where $\cos x \neq 0$ .

Read the theoryUBC Math 100 §1.5 Familiar functionsUBC Math 100 §4.6 Inverse functionsCLP-1 §0.4 FunctionsCLP-1 §0.5 Parsing formulasCLP-1 §0.6 Inverse functions

Worked example

Standard2 marks · 4 min

Let $f(x) = 3e^{2x} - 1$ . Find $f^{-1}(5)$ .

Strategy

The value of an inverse at 5 is the solution of f(x) = 5. Isolate the exponential, undo it with the natural logarithm, then divide by the coefficient of x.

Working

$3e^{2x} - 1 = 5$
f⁻¹(5) = x exactly when f(x) = 5.
$e^{2x} = 2$
Add 1, then divide by 3, so the exponential stands alone.
$2x = \log 2 \quad\Longrightarrow\quad x = \frac{\log 2}{2}$
Take log of both sides: log(eᵘ) = u.
$f\!\left(\tfrac{\log 2}{2}\right) = 3e^{\log 2} - 1 = 3(2) - 1 = 5$
Check by substituting back.

Answer

f^{-1}(5) = \dfrac{\log 2}{2} \approx 0.347

Check

f(0) = 2 and f(0.5) = 3e − 1 ≈ 7.15, so the input giving 5 lies between 0 and 0.5, as 0.347 does.

Where this goes wrong

Taking logarithms before the exponential stands alone, or forgetting to divide by 2 at the end. Get e^2x by itself first; then log turns the exponent 2x into an ordinary expression to solve.

On the paper

To find $f^{-1}(b)$ , solve $f(x) = b$ one step at a time, then check by putting your answer back into $f$ .

Your turn

Standard2 marks · 3 min

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

Answer this question

Practice questions

5 of the 19, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory2 marks · 3 min
Let $f(x) = 2x - 5$ and $g(x) = x^{2} + 1$ . Evaluate the composition below.
$(f \circ g)(3)$
Strategy
Composition is evaluated inside first. Find the output of the inner function at 3, then use that output as the input to the outer function.
Working
$(f \circ g)(3) = f\bigl(g(3)\bigr)$
Unpack the notation. The right-hand function acts first.
$g(3) = 3^{2} + 1 = 10$
Evaluate the inner function at the given input.
$f(10) = 2(10) - 5 = 15$
Substitute that single value into the outer function.
Answer
$(f \circ g)(3) = 15$
The answer is 15.
Check
As one formula, f(g(x)) = 2(x² + 1) − 5 = 2x² − 3, which at x = 3 is 18 − 3 = 15.
Where this goes wrong
Composing in the wrong order and computing g(f(3)) = g(1) = 2. The function written on the left of the circle is applied last, not first.
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The graph of $y = f(x)$ is shifted 3 units to the right and then reflected in the $x$ -axis. Which formula describes the resulting graph?
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Introductory1 mark · 2 min
Find the largest value of $x$ in the domain of the function below.
$g(x) = \frac{\sqrt{7 - 2x}}{x + 3}$
Strategy
The domain is every x for which the formula makes sense. The square root needs a non-negative argument and the denominator must not be zero; the largest allowed x comes from the square root.
Working
$7 - 2x \geq 0 \iff x \leq \tfrac{7}{2}$
The square root is defined only for a non-negative argument.
$x + 3 \neq 0 \iff x \neq -3$
The denominator excludes one more point, well to the left.
$\text{Domain: } (-\infty, -3) \cup (-3, \tfrac{7}{2}]$
The right-hand end is included: at x = 7/2 the square root is 0, which is allowed.
Answer
$x = \frac{7}{2}$
The answer is seven halves, or 3.5.
Check
g(3.5) = √0/6.5 = 0 is defined, while at x = 3.6 the root would be √(−0.2), which is not.
Where this goes wrong
Excluding 7/2 because “the square root of zero is a problem”. Zero under a square root is fine; zero in a denominator is not, and the denominator is x + 3, not the root.
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Let $f(x) = \log\!\left(\dfrac{1 + x}{1 - x}\right)$ . Find the domain of $f$ , show that $f$ is odd, and find a formula for $f^{-1}(y)$ .
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Standard2 marks · 4 min
Find the sum of all real solutions of the equation below.
$2^{2x} - 6\cdot 2^{x} + 8 = 0$
Strategy
Recognise the equation as a quadratic in 2ˣ. Substitute, factor, and then undo the substitution with logarithms, or by recognising powers of 2.
Working
$u = 2^{x}: \quad u^{2} - 6u + 8 = 0$
Because 2^2x = (2^x)², the equation is quadratic in u.
$(u - 2)(u - 4) = 0 \quad\Longrightarrow\quad u = 2 \text{ or } u = 4$
Both roots are positive, so both are attainable values of 2^x.
$2^{x} = 2 \Rightarrow x = 1, \qquad 2^{x} = 4 \Rightarrow x = 2$
A negative u would have to be discarded; here neither is.
$1 + 2 = 3$
Sum the solutions.
Answer
$x = 1 \text{ or } x = 2; \quad \text{sum } 3$
The answer is the solutions are 1 and 2, whose sum is 3.
Check
x = 1 gives 4 − 12 + 8 = 0 and x = 2 gives 16 − 24 + 8 = 0, so both are solutions.
Where this goes wrong
Stopping at u = 2 and u = 4 and adding them to get 6. Those are values of 2ˣ, not of x; each one still has to be converted back.
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Find the sum of all solutions of the equation below in the interval $[0, 2\pi)$ . $2\sin^{2} x - \sin x - 1 = 0$
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Challenging4 marks · 7 min
Let $f(x) = \log\!\left(\dfrac{1 + x}{1 - x}\right)$ . Find the domain of $f$ , show that $f$ is odd, and find a formula for $f^{-1}(y)$ .
Strategy
The three parts use three different tools: an inequality for the domain, a log identity for oddness, and exponentiation for the inverse. Do them in that order; the domain tells you what the inverse’s range must be.
Working
$\frac{1 + x}{1 - x} > 0 \iff -1 < x < 1$
Numerator and denominator must have the same sign. Both are positive exactly on (−1, 1); both negative would need x < −1 and x > 1 at once, which never happens.
$f(-x) = \log\!\left(\frac{1 - x}{1 + x}\right) = -\log\!\left(\frac{1 + x}{1 - x}\right) = -f(x)$
Swapping x for −x flips the fraction, and log(1/u) = −log u.
$e^{y} = \frac{1 + x}{1 - x} \quad\Longrightarrow\quad e^{y} - x e^{y} = 1 + x$
Exponentiate, then clear the denominator.
$x\bigl(e^{y} + 1\bigr) = e^{y} - 1 \quad\Longrightarrow\quad f^{-1}(y) = \frac{e^{y} - 1}{e^{y} + 1}$
Collect the x terms. The result lies in (−1, 1) for every y, matching the domain of f.
Answer
$\text{Domain } (-1, 1);\quad f \text{ odd};\quad f^{-1}(y) = \frac{e^{y} - 1}{e^{y} + 1}$
The answer is domain the open interval from −1 to 1; f is odd; the inverse is (e to the y minus 1) over (e to the y plus 1).
What earns the marks
Domain: (1 + x)/(1 − x) > 0, giving −1 < x < 1, with both endpoints excluded.
Oddness: f(−x) = log((1 − x)/(1 + x)) = −log((1 + x)/(1 − x)) = −f(x), using log(1/u) = −log u.
Inverse: solve e^y = (1 + x)/(1 − x) for x to get f⁻¹(y) = (e^y − 1)/(e^y + 1), defined for every real y.
Check
At y = log 3 the inverse gives (3 − 1)/(3 + 1) = 1/2, and f(1/2) = log(1.5/0.5) = log 3, so the two undo each other.
Where this goes wrong
Giving the domain as x ≠ 1, forgetting that the argument of a logarithm must be positive, not merely defined. For x < −1 the fraction is negative and f does not exist.
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Let $f(x) = \dfrac{3x - 1}{x + 2}$ . Find $f^{-1}(1)$ .
Answer this question
Capstone5 marks · 8 min
A function $f$ satisfies the identity below for every $x \neq 2$ . Find $f^{-1}(3)$ .
$f(3x - 1) = \frac{2x + 5}{x - 2}$
Strategy
You are never given f directly, only f evaluated at 3x − 1. Find the x for which that output is 3; the corresponding input to f, which is 3x − 1, is the answer.
Working
$\frac{2x + 5}{x - 2} = 3$
Find where the known outputs of f equal 3.
$2x + 5 = 3x - 6 \quad\Longrightarrow\quad x = 11$
Clear the denominator and solve.
$f(3\cdot 11 - 1) = f(32) = 3$
At x = 11 the identity says f(32) = 3.
$f^{-1}(3) = 32$
f sends 32 to 3, so f⁻¹ sends 3 back to 32.
Answer
$f^{-1}(3) = 32$
The answer is 32.
Check
Writing u = 3x − 1 gives f(u) = (2u + 17)/(u − 5), and f(32) = 81/27 = 3.
Where this goes wrong
Answering 11. Solving the equation gives the x in the identity, but the argument of f there is 3x − 1 = 32, and that is the value f⁻¹ returns.
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Let $f(x) = x^{3} + x$ . Find $f^{-1}(10)$ without finding a formula for $f^{-1}$ .
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Leads to
Limits
Limit laws, the indeterminate form 0/0, one-sided and infinite limits, limits at infinity and asymptotes.
30 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Your turn

Practice questions

Strategy

Working

Answer

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Strategy

Working

Answer

Check

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Strategy

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Check

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Strategy

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What earns the marks

Check

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Strategy

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Answer

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Related topics

Limits