Inverse Functions — Mathslive.ie

Section 1 of 5

What an inverse is

$f^{-1}(x)$ reads "$f$ inverse of $x$". It undoes what $f$ does.

From indices, $x^{-1} = \dfrac{1}{x}$. A function inverse is a different idea.

$f^{-1}(x)$ is not the same as $\dfrac{1}{f(x)}$.

(i) Couples — $f(x) = 2x+1$

Write the couples for $f(x) = 2x+1$ on the domain $-1 \le x \le 2$.

$f(x): \quad (-1,-1) \quad (0,1) \quad (1,3) \quad (2,5)$

To get $f^{-1}$, swap each pair — the $x$ and the $y$ trade places.

$f^{-1}(x): \quad (-1,-1) \quad (1,0) \quad (3,1) \quad (5,2)$

Reverse every arrow and you have $f^{-1}$. Only a bijective function can have an inverse — each output comes from exactly one input.

YOU TRY · 1

The point $(2,4)$ lies on $f$. Write the matching couple on $f^{-1}$.

Swap the $x$ and the $y$.

Swap: $(4,2)$.

$(4,2)$

YOU TRY · 2

$f$ has the couples $(1,5),\,(2,7),\,(3,9)$. Write the couples of $f^{-1}$.

Swap each pair.

$(5,1),\,(7,2),\,(9,3)$

Section 2 of 5

Inverse on the graph

$f(x)$ means: give the $x$ value, find the $y$.

$f^{-1}(x)$ means: give the $y$ value, find the $x$.

Everyday version: I pay you €$10$ an hour. "How much for 5 hrs?" is $f(x)$. "How long to earn €$80$?" is $f^{-1}(x)$.

Plot the couples of $f$ and $f^{-1}$ on the same axes:

Key fact

•$f^{-1}(x)$ is the image of $f(x)$ under axial symmetry in the line $y = x$.

YOU TRY · 3

The point $(3,7)$ lies on $f$. State the point that must lie on $f^{-1}$.

Reflect in $y = x$ — swap the coordinates.

$(7,3)$

YOU TRY · 4

A function and its inverse are reflections of each other in which line?

The line of axial symmetry.

$y = x$

Section 3 of 5

Finding the inverse — linear

Method: write $y = f(x)$, then make $x$ the subject. That swapped rule is $f^{-1}$.

(i) $f(x) = 2x+1$

$y = 2x+1$

$y - 1 = 2x$

$\dfrac{y-1}{2} = x$

$f^{-1}(x) = \dfrac{x-1}{2}$

(ii) $g(x) = \dfrac{5x-1}{3}, \quad x \in \mathbb{R}$

$y = \dfrac{5x-1}{3}$

$3y = 5x - 1$

$3y + 1 = 5x$

$\dfrac{3y+1}{5} = x$

$g^{-1}(x) = \dfrac{3x+1}{5}$

(iii) $f(x) = \dfrac{3x-1}{11}$

$y = \dfrac{3x-1}{11}$

$11y = 3x - 1$

$11y + 1 = 3x$

$f^{-1}(x) = \dfrac{11x+1}{3}$

YOU TRY · 5

$f(x) = 4x + 3$. Find $f^{-1}(x)$.

$y = 4x+3$, then make $x$ the subject.

$y - 3 = 4x \;\Rightarrow\; \dfrac{y-3}{4} = x$

$f^{-1}(x) = \dfrac{x-3}{4}$

YOU TRY · 6

$g(x) = \dfrac{2x+5}{7}$. Find $g^{-1}(x)$.

Multiply across by 7 first.

$7y = 2x+5 \;\Rightarrow\; 7y-5 = 2x$

$g^{-1}(x) = \dfrac{7x-5}{2}$

YOU TRY · 7

$f(x) = 6 - x$. Find $f^{-1}(x)$.

$y = 6 - x$. Solve for $x$.

$y = 6 - x \;\Rightarrow\; x = 6 - y$

$f^{-1}(x) = 6 - x$ (self-inverse)

$f^{-1}(x) = 6 - x$

Section 4 of 5

Inverses of quadratics

A quadratic isn't one-to-one until we restrict the domain to one side of the vertex. Then complete the square to invert.

(i) $f(x) = x^2 - 6x + 1, \quad x \ge 3$

First instinct — factor out $x$:

$y - 1 = x(x-6)$ (stalls — $x$ sits in both factors)

Complete the square instead:

$y = x^2 - 6x + 9 + 1 - 9$

$y = (x-3)^2 - 8$ vertex $(3,-8)$

$y + 8 = (x-3)^2$

$\sqrt{y+8} = x - 3$

$\sqrt{y+8} + 3 = x$

$f^{-1}(x) = \sqrt{x+8} + 3$

Domain of $f$ is $x \ge 3$, so the range is $y \ge -8$. Hence:

Domain of $f^{-1}$: $x \ge -8$

(ii) $g(x) = x^2 + 8x - 5$

$y = x^2 + 8x + 16 - 5 - 16$

$y = (x+4)^2 - 21$ vertex $(-4,-21)$, domain $x \ge -4$

$y + 21 = (x+4)^2$

$\sqrt{y+21} = x + 4$

$\sqrt{y+21} - 4 = x$

$g^{-1}(x) = \sqrt{x+21} - 4$

YOU TRY · 8

$f(x) = x^2 - 4x + 1, \; x \ge 2$. Find $f^{-1}(x)$.

Complete the square, then solve for $x$.

$y = (x-2)^2 - 3 \;\Rightarrow\; \sqrt{y+3} + 2 = x$

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

YOU TRY · 9

$g(x) = x^2 + 10x, \; x \ge -5$. Find $g^{-1}(x)$.

$x^2 + 10x = (x+5)^2 - 25$.

$y = (x+5)^2 - 25 \;\Rightarrow\; \sqrt{y+25} - 5 = x$

$g^{-1}(x) = \sqrt{x+25} - 5$

Section 5 of 5

Recap — must learn

Must learn

1.$f^{-1}$ swaps $x$ and $y$ — give the $y$, find the $x$.

2.$f^{-1}$ is the image of $f$ under axial symmetry in $y = x$.

3.Only a bijective function has an inverse.

4.To find it: write $y = f(x)$, make $x$ the subject, then write $f^{-1}(x)$.

5.$f^{-1}(x)$ is not $\dfrac{1}{f(x)}$.

6.Quadratics: restrict the domain, complete the square, then solve.

End of lesson

Inverse Functions — HL · Mathslive.ie