A function is like an equation that relates an input to an output. We commonly express functions in the form:

$f(x)$f(x)$=$=...

A linear function is a relationship between two variables that, when graphed, will be in a straight line.

An inverse function is a function that reverses another function. It sounds a bit confusing so let's look at how to find inverses in a bit more detail.

Finding the Inverse of Coordinates

The inverse of a pair of coordinates (a point) can be thought of as the corresponding point if the point was reflected along the line $y=x$y=x.

For example, the line $y=x$y=x has been graphed and I have plotted the point $\left(-2,3\right)$(−2,3) in blue.

So to find the inverse, I reflect across this line. To do this I reverse the coordinates. This mean that the $y$y becomes the $x$x and the $x$x becomes the $y$y. So $\left(-2,3\right)$(−2,3) becomes $\left(3,-2\right)$(3,−2).

This is shown on the graph in green.

The reflection is evident by the fact that I have crossed the line $y=x$y=x at $90^\circ$90°.

Finding the Inverse of a Function

We'll run through the process of finding an inverse function it using an example: $f(x)$f(x)$=$=$3x+6$3x+6

1. Start with the original function, substituting $y$y for $f(x)$f(x):

$y=3x+6$y=3x+6

2. Rearrange the equation to make $x$x the subject:

$y$y

$=$=

$3x+6$3x+6

$y-6$y−6

$=$=

$3x$3x

$\frac{y-6}{3}$y−63

$=$=

$x$x

$x$x

$=$=

$\frac{y-6}{3}$y−63

3. Switch the places of $x$x and $y$y in the equation you found in step 2:

We started with $x=\frac{y-6}{3}$x=y−63, so when we switch the variables to make it $y=\frac{x-6}{3}$y=x−63

So the inverse function of $f(x)$f(x)$=$=$3x+6$3x+6 is $f(x)$f(x)$=$=$\frac{x-6}{3}$x−63

Now let's look at some examples.

Worked Examples

Question 1

Question 2

Define $y$y, the inverse of $f\left(x\right)=-\frac{1}{8}x+10$f(x)=−18x+10.

Question 3

Find the inverse of $3x+48=8y$3x+48=8y.

Question 4

If $f\left(x\right)$f(x)$=$=$kx$kx$-$−$7$7 and $f^{-1}\left(x\right)$f−1(x)$=$=$2x$2x$+$+$14$14, solve for the value of $k$k.