2.10 Composite and inverse functions — Practise

Work through these on your own. For a composite, work from the inside out. For an inverse, make the input the subject and then swap the letters; finish by composing both ways to check. State a domain restriction when the function is not one-to-one on the whole real line.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Given $f(x) = 2x + 3$ and $g(x) = x^2$ , find $f(g(x))$ .
Given $f(x) = 2x + 3$ and $g(x) = x^2$ , find $g(f(x))$ .
Given $f(x) = x^2 + 1$ and $g(x) = 3x - 1$ , work out $f(g(2))$ .
Find the inverse of $f(x) = 3x - 2$ .
Find the inverse of $f(x) = \dfrac{x + 1}{x - 2}$ , where $x \neq 2$ .
The function $f(x) = x^2 + 1$ has domain $x \geq 0$ . Find its inverse.
Find the inverse of $f(x) = x^3 + 2$ .
Given $f(x) = 2x - 1$ , find $f(f(x))$ .
For $f(x) = \sqrt{x - 2}$ , state the domain and the range, and work out $f(6)$ .
Explain why $f(x) = x^2$ has no inverse function on the whole real line.
Show that $f(x) = 5x + 1$ and $g(x) = \dfrac{x - 1}{5}$ are inverses of each other.
Given $f(x) = x^2 + 3$ on $x \geq 0$ , show that $g(x) = \sqrt{x - 3}$ is its inverse.

Answers and concise solutions

1. The composite function

Answer: $f(g(x))$ means $f$ applied to $g(x)$ : substitute $x^2$ into $f$ . So $f(g(x)) = 2x^2 + 3$ . The output of $g$ becomes the input of $f$ .

2. The composite function

Answer: $g(f(x))$ substitutes $2x + 3$ into $g$ : $g(f(x)) = (2x + 3)^2$ . Notice that $f(g(x)) = 2x^2 + 3$ is not the same as $g(f(x)) = (2x + 3)^2$ , so the order of composition matters.

3. The value

Answer: Work from the inside out. First $g(2) = 3 \times 2 - 1 = 5$ . Then $f(5) = 5^2 + 1 = 26$ . So $f(g(2)) = 26$ .

4. The inverse function

Answer: Write $y = 3x - 2$ and make $x$ the subject: $y + 2 = 3x$ , so $x = \frac{y + 2}{3}$ . Swap the letters: $f^{-1}(x) = \frac{x + 2}{3}$ . Check both ways: $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$ .

5. The inverse function

Answer: Write $y = \frac{x + 1}{x - 2}$ and cross-multiply: $y(x - 2) = x + 1$ , so $yx - 2y = x + 1$ . Collect the $x$ terms: $x(y - 1) = 2y + 1$ , so $x = \frac{2y + 1}{y - 1}$ . Swap the letters: $f^{-1}(x) = \frac{2x + 1}{x - 1}$ , where $x \neq 1$ .

6. The inverse function

Answer: Write $y = x^2 + 1$ , so $x^2 = y - 1$ and $x = \pm\sqrt{y - 1}$ . The domain $x \geq 0$ selects the positive root, so $x = \sqrt{y - 1}$ . Swap: $f^{-1}(x) = \sqrt{x - 1}$ with domain $x \geq 1$ and range $f^{-1}(x) \geq 0$ .

7. The inverse function

Answer: Write $y = x^3 + 2$ , so $x^3 = y - 2$ and $x = \sqrt[3]{y - 2}$ . Swap the letters: $f^{-1}(x) = \sqrt[3]{x - 2}$ . A cube has one real root for every input, so no restriction on the domain is needed.

8. The composite function

Answer: Substitute $f(x) = 2x - 1$ into $f$ : $f(f(x)) = 2(2x - 1) - 1 = 4x - 2 - 1 = 4x - 3$ . Applying a function twice is written $f^2(x)$ .

9. The domain, the range and a value

Answer: The expression under the square root must not be negative, so $x - 2 \geq 0$ and the domain is $x \geq 2$ . A square root is never negative, so the range is $f(x) \geq 0$ . For $f(6)$ : $\sqrt{6 - 2} = \sqrt{4} = 2$ . At $x = 1$ the input inside the root is $-1$ , which has no real square root, so $x = 1$ is outside the domain.

10. A reason, with two inputs

Answer: An inverse needs each output to come from one input. But $f(2) = 4$ and $f(-2) = 4$ : two different inputs give the same output, so reversing the function would not know whether to return $2$ or $-2$ . The function is not one-to-one. Restricting the domain to $x \geq 0$ makes it one-to-one, and an inverse then exists.

11. Both compositions

Answer: Compose in both orders. $f(g(x)) = 5\left(\frac{x - 1}{5}\right) + 1 = (x - 1) + 1 = x$ . And $g(f(x)) = \frac{(5x + 1) - 1}{5} = \frac{5x}{5} = x$ . Both compositions return $x$ , so each function undoes the other.

12. Both compositions

Answer: $g(f(x)) = \sqrt{(x^2 + 3) - 3} = \sqrt{x^2}$ . Because the domain of $f$ is $x \geq 0$ , the value of $x$ is not negative, so $\sqrt{x^2} = x$ . In the other direction, $f(g(x)) = \left(\sqrt{x - 3}\right)^2 + 3 = (x - 3) + 3 = x$ for $x \geq 3$ . Both compositions return $x$ , so $g$ is the inverse of $f$ .