Inverse Functions

Q: Is f inverse the same as 1 over f?

No. $f^{-1}$ is the function that undoes $f$. $\frac{1}{f(x)}$ is the reciprocal, which is something completely different.

Q: Why do we restrict the domain of a parabola before finding its inverse?

Without a restriction, the inverse gives two $y$-values for each $x$, so it is not a function. Restricting to one side of the turning point fixes this.

An inverse function undoes what the original function does, and its graph is the reflection of the original in the line $y = x$ . IEB Paper 1 tests inverses of linear functions, of $y = a x^{2}$ with a restricted domain, and of exponential functions through logarithms.

Inverse of a linear function

To find the inverse of $y = f (x)$ :

Swap $x$ and $y$ .
Make $y$ the subject.
Write the answer as $f^{- 1} (x) = \dots$

Example. $f (x) = 3 x - 6$

$x = 3 y - 6 \Rightarrow y = \frac{x + 6}{3}$

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

Check: $f (f^{- 1} (x)) = 3 (\frac{1}{3} x + 2) - 6 = x$ . The inverse of a non-horizontal straight line is always a function.

Inverse of y = ax² with a restricted domain

For $f (x) = a x^{2}$ , swapping gives $x = a y^{2}$ , so $y = \pm \frac{x}{a}$ . With both signs this is not a function, because each $x$ gives two $y$ -values.

Restrict the domain of $f$ to make the inverse a function:

If $f$ has domain $x \geq 0$ , then $f^{- 1} (x) = \frac{x}{a}$ .
If $f$ has domain $x \leq 0$ , then $f^{- 1} (x) = - \frac{x}{a}$ .

Example. $g (x) = 2 x^{2}$ , $x \geq 0$ . Then $x = 2 y^{2}$ , so $g^{- 1} (x) = \frac{x}{2}$ with domain $x \geq 0$ .

The domain of $f$ becomes the range of $f^{- 1}$ , and the range of $f$ becomes the domain of $f^{- 1}$ .

The relation x = f(y)

Swapping $x$ and $y$ in $y = f (x)$ gives $x = f (y)$ . This is the inverse relation, whether or not it is a function. Its graph is the reflection of $y = f (x)$ in the line $y = x$ .

Example. The inverse relation of $y = x^{2}$ is $x = y^{2}$ , a parabola lying on its side. It fails the vertical line test.

To sketch an inverse, reflect key points: $(a; b)$ on $f$ becomes $(b; a)$ on $f^{- 1}$ . Points on the line $y = x$ stay where they are, which is why $f$ and $f^{- 1}$ often meet on that line.

When the inverse is a function

The inverse of $f$ is a function only if $f$ is one-to-one: every $y$ -value comes from exactly one $x$ -value. Use the horizontal line test on the graph of $f$ : if any horizontal line cuts $f$ more than once, $f^{- 1}$ is not a function.

Linear functions (not horizontal) and exponential functions are one-to-one.
Parabolas are many-to-one, so their domain must be restricted.

Example. $y = (x - 1)^{2}$ fails the horizontal line test, because $y = 4$ cuts it at $x = - 1$ and $x = 3$ . Restricting to $x \geq 1$ makes it one-to-one.

Key terms

Inverse function: A function $f^{- 1}$ that reverses $f$ , so $f^{- 1} (f (x)) = x$ .
One-to-one: Each output comes from exactly one input.
Many-to-one: Some outputs come from more than one input, such as a parabola.
Horizontal line test: If a horizontal line cuts the graph of $f$ more than once, the inverse is not a function.
Restricted domain: A limited set of $x$ -values chosen so that $f$ becomes one-to-one.

Common mistakes

Writing $f^{- 1} (x)$ as $\frac{1}{f ( x )}$ . The $- 1$ is not an exponent.
Forgetting the negative square root when the restricted domain is $x \leq 0$ .
Not swapping the domain and range when stating them for the inverse.
Reflecting in the $x$ -axis or $y$ -axis instead of the line $y = x$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Given $f (x) = 3 x - 6$ . Determine $f^{- 1} (x)$ in the form $f^{- 1} (x) = \dots$

Worked solution

Hidden - try the question first

Solution

$x = 3 y - 6$
$3 y = x + 6$
$f^{- 1} (x) = \frac{1}{3} x + 2$

Question 2

5 marks total

routine

Given $g (x) = 2 x^{2}$ , $x \geq 0$ .

(a) Determine $g^{- 1} (x)$ . (3)

(b) Write down the domain and range of $g^{- 1}$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $x = 2 y^{2}$
$y^{2} = \frac{x}{2}$
$g^{- 1} (x) = \frac{x}{2}$ (positive root, since the range of $g^{- 1}$ is $y \geq 0$ )
(b) Domain: $x \geq 0$
Range: $y \geq 0$

Question 3

6 marks total

complex

$f (x) = a x^{2}$ , $x \leq 0$ , passes through the point $(- 4; - 8)$ .

(a) Determine the value of $a$ . (1)

(b) Determine $f^{- 1} (x)$ and state its domain. (3)

Worked solution

Hidden - try the question first

Solution

(a) $x = 2 y^{2}$
(b) $x = - \frac{1}{2} y^{2} \Rightarrow y^{2} = - 2 x$
$f^{- 1} (x) = - - 2 x$ , taking the negative root because the domain of $f$ is $x \leq 0$
Domain of $f^{- 1}$ : $x \leq 0$
(c) They meet on $y = x$ : $- \frac{1}{2} x^{2} = x \Rightarrow x (x + 2) = 0$
$(0; 0)$ and $(- 2; - 2)$

Question 4

8 marks total

problem-solving

The straight line $f$ passes through $(- 2; 0)$ and $(0; 4)$ .

(a) Determine the equations of $f$ and $f^{- 1}$ . (3)

(b) Determine the point where $f$ and $f^{- 1}$ intersect. (2)

(d) $g (x) = f^{- 1} (x) + k$ passes through $(2; 3)$ . Determine $k$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $x = 2 y^{2}$ , so $f (x) = 2 x + 4$
$x = 2 y + 4$
$f^{- 1} (x) = \frac{1}{2} x - 2$
(b) $2 x + 4 = \frac{1}{2} x - 2 \Rightarrow \frac{3}{2} x = - 6 \Rightarrow x = - 4$
$(- 4; - 4)$
(c) $x > - 4$
(d) $3 = \frac{1}{2} (2) - 2 + k$
$k = 4$

FAQs

Is f inverse the same as 1 over f?

No. $f^{- 1}$ is the function that undoes $f$ . $\frac{1}{f ( x )}$ is the reciprocal, which is something completely different.

Why do we restrict the domain of a parabola before finding its inverse?

Without a restriction, the inverse gives two $y$ -values for each $x$ , so it is not a function. Restricting to one side of the turning point fixes this.

How do I sketch an inverse quickly?

Swap the coordinates of a few key points and reflect the shape in the line $y = x$ . Intercepts on one axis become intercepts on the other.

Inverse of a linear function

Inverse of y = ax² with a restricted domain

The relation x = f(y)

When the inverse is a function

Key terms

Common mistakes

Practice questions

FAQs

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