Week of December 2, 2019

Finding inverse functions

Inverse functions, in the most general sense, are functions that "reverse" each other. For example, if $f$f takes $a$a to $b$b, then the inverse, $f^{-1}$f, start superscript, minus, 1, end superscript, must take $b$b to $a$a.

$\blueD{a}$$f$$\goldD{b}$$f^{-1}$$\blueD{a}$

Or in other words, $f(a)=b \iff f^{-1}(b)=a$f, left parenthesis, a, right parenthesis, equals, b, \Longleftrightarrow, f, start superscript, minus, 1, end superscript, left parenthesis, b, right parenthesis, equals, a.

In this article we will learn how to find the formula of the inverse function when we have the formula of the original function.

Before we start...

In this lesson, we will find the inverse function of $f(x)=3x+2$f, left parenthesis, x, right parenthesis, equals, 3, x, plus, 2.

Before we do that, let's first think about how we would find $f^{-1}(8)$f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis.

To find $f^{-1}(8)$f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis, we need to find the input of $f$f that corresponds to an output of $8$8. This is because if $f^{-1}(8)=x$f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis, equals, x, then by definition of inverses, $f(x)=8$f, left parenthesis, x, right parenthesis, equals, 8.

$\begin{aligned} f(x) &= 3 x+2\\\\ 8 &= 3 x+2 &&\small{\gray{\text{Let f(x)=8}}} \\\\6&=3x &&\small{\gray{\text{Subtract 2 from both sides}}}\\\\ 2&=x &&\small{\gray{\text{Divide both sides by 3}}} \end{aligned}$

So $f(2)=8$f, left parenthesis, 2, right parenthesis, equals, 8 which means that $f^{-1}(8)=2$f, start superscript, minus, 1, end superscript, left parenthesis, 8, right parenthesis, equals, 2

Finding inverse functions

We can generalize what we did above to find $f^{-1}(y)$f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis for any $y$y. 

[Why did we use y here?]

$x$x

$x$x

To find $f^{-1}(y)$f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis, we can find the input of $f$f that corresponds to an output of $y$y. This is because if $f^{-1}(y)=x$f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis, equals, x then by definition of inverses, $f(x)=y$f, left parenthesis, x, right parenthesis, equals, y.

$\begin{aligned} f(x) &= 3 x+2\\\\ y &= 3 x+2 &&\small{\gray{\text{Let f(x)=y}}} \\\\y-2&=3x &&\small{\gray{\text{Subtract 2 from both sides}}}\\\\ \dfrac{y-2}{3}&=x &&\small{\gray{\text{Divide both sides by 3}}} \end{aligned}$

So $f^{-1}(y)=\dfrac{y-2}{3}$f, start superscript, minus, 1, end superscript, left parenthesis, y, right parenthesis, equals, start fraction, y, minus, 2, divided by, 3, end fraction.

Since the choice of the variable is arbitrary, we can write this as $f^{-1}(x)=\dfrac{x-2}{3}$f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 2, divided by, 3, end fraction.

[Is there another way to do this?]

$f^{-1}$f, start superscript, minus, 1, end superscript$x$x$y$y$y$y

$\begin{aligned} f(x)&=3x+2\\\\ \goldD y &= 3\blueD x+2 &&\small{\gray{\text{Replace f(x) with y}}}\\\\ \blueD x &= 3\goldD y+2 &&\small{\gray{\text{Switch x and y}}} \\\\x-2&=3y &&\small{\gray{\text{Subtract 2 from both sides}}}\\\\ \dfrac{x-2}{3}&=y &&\small{\gray{\text{Divide both sides by 3}}} \\\\ \end{aligned}$

$f^{-1}(x)=\dfrac{x-2}{3}$f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 2, divided by, 3, end fraction

Check your understanding

1) Linear function

Find the inverse of $g(x)=2x-5$g, left parenthesis, x, right parenthesis, equals, 2, x, minus, 5.

$g^{-1}(x)=$g, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals 

[I need help!]

$g(x)$g, left parenthesis, x, right parenthesis$y$y$x$x

$\begin{aligned} g(x)&= 2x-5 \\\\ y &= 2x-5 &&\small{\gray{\text{Replace g(x) with y}}}\\\\ y+5 &= 2x &&\small{\gray{\text{Add 5 to both sides}}}\\\\ \dfrac{y+5}{2}&=x &&\small{\gray{\text{Divide both sides by 2}}} \\\\ \dfrac{y+5}{2}&=g^{-1}(y) &&\small{\gray{\text{Replace x with }g^{-1}(y)}} \end{aligned}$

$y$y$x$x$x$x

$g^{-1}(x)=\dfrac{x+5}{2}$g, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, start fraction, x, plus, 5, divided by, 2, end fraction

2) Cubic function

Find the inverse of $h(x)=x^3+2$h, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 2.

$h^{-1}(x)=$h, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals 

[I need help!]

$h(x)$h, left parenthesis, x, right parenthesis$y$y$x$x

h(x)yy−2y−2−−−−√3y−2−−−−√3=x3+2=x3+2=x3=x=h−1(y)Replace h(x) with ySubtract 2 from both sidesCube root of both sidesReplace x with h^{-1}(y)

$y$y$x$x$x$x

$h^{-1}(x)=\sqrt[3]{x-2}$h, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, cube root of, x, minus, 2, end cube root

3) Cube-root function

Find the inverse of $f(x)=4\cdot \sqrt[\Large3]{x}$f, left parenthesis, x, right parenthesis, equals, 4, dot, cube root of, x, end cube root.

$f^{-1}(x)=$f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals 

[I need help!]

$f(x)$f, left parenthesis, x, right parenthesis$y$y$x$x

$\begin{aligned} f(x)&= 4\sqrt[3]{x} \\\\ y &= 4\sqrt[3]{x} &&\small{\gray{\text{Replace f(x) with y}}}\\\\ \dfrac{y}{4} &=\sqrt[3]{x}&&\small{\gray{\text{Divide both sides by 4}}}\\\\ \left(\dfrac{y}{4}\right)^3&=x&&\small{\gray{\text{Cube both sides}}} \\\\ \left(\dfrac{y}{4}\right)^3&=f^{-1}(y)&&\small{\gray{\text{Replace x with }f^{-1}(y)}} \\\\ \end{aligned}$

$y$y$x$x$x$x

$f^{-1}(x)=\left(\dfrac{x}{4}\right)^3$f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, left parenthesis, start fraction, x, divided by, 4, end fraction, right parenthesis, cubed$f^{-1}(x)=\dfrac{x^3}{64}$f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, start fraction, x, cubed, divided by, 64, end fraction

4) Rational functions

Find the inverse of $g(x)=\dfrac{x-3}{x-2}$g, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 3, divided by, x, minus, 2, end fraction.

$g^{-1}(x)=$g, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals 

[I need help!]

$g(x)$g, left parenthesis, x, right parenthesis$y$y$x$x

$\begin{aligned} g(x)&= \dfrac{x-3}{x-2} \\\\\\ y&= \dfrac{x-3}{x-2} &&\small{\gray{\text{Replace g(x) with y}}}\\\\\\ y(x-2) &= x-3 &&\small{\gray{\text{Multiply both sides by x-2}}}\\\\ yx-2y&= x-3 &&\small{\gray{\text{Distribute}}} \\\\ yx-x&=2y-3&&\small{\gray{\text{Group terms with x on one side}}}\\\\ x(y-1)&=2y-3&&\small{\gray{\text{Factor out an x}}}\\\\ x&=\dfrac{2y-3}{y-1} &&\small{\gray{\text{Divide both sides by y-1}}}\\\\ g^{-1}(y)&=\dfrac{2y-3}{y-1} &&\small{\gray{\text{Replace x with }g^{-1}(y)}}\end{aligned}$

$y$y$x$x$x$x

$g^{-1}(x)=\dfrac{2x-3}{x-1}$g, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis, equals, start fraction, 2, x, minus, 3, divided by, x, minus, 1, end fraction

5) Challenge problem

Match each function with the type of its inverse.

Function

Inverse type

$f(x)=3x-5$f, left parenthesis, x, right parenthesis, equals, 3, x, minus, 5 $g(x)=x^3-7$g, left parenthesis, x, right parenthesis, equals, x, cubed, minus, 7 $h(x)=\dfrac{x+2}{x-3}$h, left parenthesis, x, right parenthesis, equals, start fraction, x, plus, 2, divided by, x, minus, 3, end fraction $j(x)=\sqrt{x+2}$j, left parenthesis, x, right parenthesis, equals, square root of, x, plus, 2, end square root

Quadratic Cube root Linear Rational

Source:https://www.khanacademy.org/math/algebra-home/alg-functions/alg-finding-inverse-functions/a/finding-inverse-functions