2.10 Composite and inverse functions — Practise with help

Work each one first, closed book. For a composite, decide which function acts first. For an inverse, write $y = f(x)$ and make $x$ the subject before you swap the letters.

Warm-up: remember

Work out $f(3)$ for $f(x) = 2x + 1$ .
Make $x$ the subject of $y = 4x - 5$ .
Solve $x^2 = 9$ .

Answers

$f(3) = 2 \times 3 + 1 = 7$ .
$y + 5 = 4x$ , so $x = \dfrac{y + 5}{4}$ .
$x = 3$ or $x = -3$ .

If question 2 was hard, revisit rearranging: undo each operation in reverse order.

Practise with help

1. Given $f(x) = x + 4$ and $g(x) = 2x$ , find $f(g(x))$ and $g(f(x))$ .

Hint

For $f(g(x))$ , the input to $f$ is $2x$ . For $g(f(x))$ , the input to $g$ is $x + 4$ .

Next step

$f(g(x)) = (2x) + 4$ , and $g(f(x)) = 2(x + 4)$ .

Full working

For $f(g(x))$ , substitute $g(x) = 2x$ into $f$ :

f(g(x)) = 2x + 4.

For $g(f(x))$ , substitute $f(x) = x + 4$ into $g$ :

g(f(x)) = 2(x + 4) = 2x + 8.

The two results differ, so the order matters.

2. Given $f(x) = x^2 - 1$ and $g(x) = x + 2$ , find $f(g(1))$ .

Hint

Work from the inside out: find $g(1)$ first.

Next step

$g(1) = 3$ , so work out $f(3)$ .

Full working

First $g(1) = 1 + 2 = 3$ . Then

f(3) = 3^2 - 1 = 8.

So $f(g(1)) = 8$ .

3. Find the inverse of $f(x) = 4x + 7$ .

Hint

Write $y = 4x + 7$ , then make $x$ the subject by subtracting $7$ and dividing by $4$ .

Next step

$y - 7 = 4x$ , so $x = \dfrac{y - 7}{4}$ . Now swap the letters.

Full working

Write $y = 4x + 7$ and rearrange:

y - 7 = 4x, \qquad x = \frac{y - 7}{4}.

Swap the letters:

f^{-1}(x) = \frac{x - 7}{4}.

Check: $f^{-1}(f(x)) = \frac{(4x + 7) - 7}{4} = \frac{4x}{4} = x$ .

4. Find the inverse of $f(x) = \dfrac{2x + 1}{x - 3}$ , where $x \neq 3$ .

Hint

Write $y = \frac{2x + 1}{x - 3}$ and multiply both sides by $x - 3$ . Then collect the $x$ terms.

Next step

$y(x - 3) = 2x + 1$ gives $yx - 3y = 2x + 1$ , so $x(y - 2) = 3y + 1$ .

Full working

Write $y = \frac{2x + 1}{x - 3}$ and cross-multiply:

y(x - 3) = 2x + 1, \qquad yx - 3y = 2x + 1.

Collect the $x$ terms:

yx - 2x = 3y + 1, \qquad x(y - 2) = 3y + 1, \qquad x = \frac{3y + 1}{y - 2}.

Swap the letters:

f^{-1}(x) = \frac{3x + 1}{x - 2}, \qquad x \neq 2.

5. Show that $f(x) = x^2$ has no inverse on the whole real line, and find the inverse of $f$ on the domain $x \geq 0$ .

Hint

For the first part, find two different inputs with the same output. For the second, take the positive square root only.

Next step

$f(3) = 9$ and $f(-3) = 9$ , so the function is not one-to-one. On $x \geq 0$ , write $y = x^2$ and take the positive root.

Full working

Two different inputs give the same output:

f(3) = 3^2 = 9, \qquad f(-3) = (-3)^2 = 9.

So the function is not one-to-one, and a reverse rule would not know whether to return $3$ or $-3$ . Restrict the domain to $x \geq 0$ . Then $y = x^2$ gives $x = \sqrt{y}$ , and swapping the letters gives

f^{-1}(x) = \sqrt{x}, \qquad x \geq 0.

6. The function $g(x) = \sqrt{x - 1}$ is defined for $x \geq 1$ . Find its domain and range, and work out $g(10)$ .

Hint

The expression under the root must not be negative, which fixes the domain. A square root is never negative, which fixes the range.

Next step

The domain is $x - 1 \geq 0$ . For the range, ask what values $\sqrt{x - 1}$ can take as $x$ runs through the domain.

Full working

The input of the square root must not be negative:

x - 1 \geq 0, \qquad x \geq 1.

So the domain is $x \geq 1$ . Since a square root is never negative, the range is $g(x) \geq 0$ . For $x = 10$ :

g(10) = \sqrt{10 - 1} = \sqrt{9} = 3.