Composite and Inverse Functions

Inverse Functions

Relating a Function with Its Inverse

Two functions ff and gg are inverse functions if f(g(x))=g(f(x))=xf(g(x)) = g(f(x)) = x. To determine the rule for an inverse function, write as y=f(x)y=f(x), reverse xx and yy, and then solve for yy. The domain of ff is the range of gg, and vice versa.

If the result of switching the inputs and outputs of a function is also a function, it is called the inverse function. The composition of a function and its inverse is the identity function.

The inverse of the function f(x)f(x) is written as f1(x)f^{-1}(x). This is read "ff inverse of xx." This notation does not mean the same thing as:
(f(x))1=1f(x)(f(x))^{-1}=\frac{1}{f(x)}
The inverse of a function is defined only if the function is one-to-one.
(ff1)(x)=(f1f)(x)=x(f \circ f^{-1})(x)=(f^{-1} \circ f)(x)=x
If the output of a function is used as the input of its inverse function, then the result is the original input. The inverse function can be said to "undo" the operations performed by the function.
To identify the inverse of a function that is given as a rule, mapping, or graph, switch the inputs and outputs, or swap xx and yy.
Step-By-Step Example
Identifying the Inverse of a Linear Function
Determine the inverse of:
f(x)=8x+5f(x)=8x+5
Step 1
Write f(x)f(x) as yy.
y=8x+5y=8x+5
Step 2
Switch the positions of xx and yy.
x=8y+5x=8y+5
Step 3
Solve for yy.
x=8y+5x5=8y+55x5=8yx58=y\begin{aligned}x&=8y+5\\x-5&=8y+5-5\\x-5&=8y\\\frac{x-5}{8}&=y\end{aligned}
Step 4
Replace yy with f1(x)f^{-1}(x).
f1(x)=x58f^{-1}(x)=\frac{x-5}{8}
Solution
The inverse of the function is:
f1(x)=x58f^{-1}(x)=\frac{x-5}{8}
Check that the composition of the function and its inverse is the identity by using the function:
f1(f(x))=xf^{-1}(f(x))=x
Calculate:
f1(f(x))=f1(8x+5)=(8x+5)58=8x8=x\begin{aligned}f^{-1}(f(x))&=f^{-1}(8x+5)\\&=\frac{(8x+5)-5}{8}\\&=\frac{8x}{8}\\&=x\end{aligned}

Step-By-Step Example
Identifying the Inverse of a Radical Function
Determine the inverse of the radical function:
f(x)=4x+6f(x)=\sqrt {4x+6}
Step 1
Write f(x)f(x) as yy.
y=4x+6y=\sqrt {4x+6}
Step 2
Switch the positions of xx and yy.
x=4y+6x=\sqrt {4y+6}
Step 3
Solve for yy.
x=4y+6(x)2=(4y+6)2x2=4y+6x26=4y+66x26=4yx264=y\begin{aligned}x&=\sqrt{4y+6}\\(x)^2&=(\sqrt{4y+6})^2\\x^2&=4y+6\\x^2-6&=4y+6-6\\x^2-6&=4y\\\frac{x^2-6}{4}&=y\end{aligned}
Step 4
Replace yy with f1(x)f^{-1}(x).
f1(x)=x264f^{-1}(x)=\frac{x^2-6}{4}
Solution
The inverse of the radical function is:
f1(x)=x264f^{-1}(x)=\frac{x^2-6}{4}
Check that composition of the function and its inverse is the identity by checking that:
f1(f(x))=xf^{-1}(f(x))=x
Calculate:
f1(f(x))=f1(4x+6)=(4x+6)264=4x+664=4x4=x\begin{aligned} f^{-1}(f(x))&=f^{-1}(\sqrt{4x+6})\\&=\frac{(\sqrt{4x+6})^2-6}{4}\\&=\frac{4x+6-6}{4}\\&=\frac{4x}{4}\\&=x\end{aligned}

Graphs of Inverse Functions

The graphs of inverse functions are reflections of each other across the line y=xy = x.
The inverse of a function given as a set of ordered pairs is obtained by switching the xx- and yy-coordinates.
f(x)={(4,3),(2,6),(8,4),(4,7),(8,1)}f(x) = \left \{ (-4,3),(2,6),(8,-4),(4,7),(-8,1)\right \}
f1(x)={(3,4),(6,2),(4,8),(7,4),(1,8)}f^{-1}(x) = \left \{(3,-4),(6,2),(-4,8),(7,4),(1,-8) \right \}
When the xx- and yy-coordinates of a function are switched, the graph is reflected (flipped) across the line y=xy=x. So the graph of an inverse function is the reflection (mirror image) of the function across the line y=xy=x.
Step-By-Step Example
Graphing the Inverse of a Function
Identify the inverse of the given function:
f(x)=62xf(x)=6-2x
Then graph both the given function and its inverse.
Step 1
Replace f(x)f(x) with yy.
y=62xy=6-2x
Step 2
Switch the positions of xx and yy.
x=62yx=6-2y
Step 3
Solve for yy.
x=62yx+2yx=62y+2yx2y=6xy=3x2\begin{aligned}x&=6-2y\\x+2y-x&=6-2y+2y-x\\2y&=6-x\\y&=3-\frac{x}{2}\end{aligned}
Step 4
Replace yy with f1(x)f^{-1}(x).
f1(x)=3x2f^{-1}(x)=3-\frac{x}{2}
Solution
The inverse is:
f1(x)=3x2f^{-1}(x)=3-\frac{x}{2}
Graph both functions. The graph of f1(x)f^{-1}(x) is the reflection of the graph of f(x)f(x) across the line y=xy=x.

Restricting the Domain

If the inverse of a function is not a function, the domain of the function can be restricted to values that produce a function for the inverse.

The inverse of a function is defined only if the function is one-to-one. One way to test whether a function is one-to-one is the horizontal line test. If the function does not pass the horizontal line test, then the inverse of the function is not a function.

Analyzing the graph of a function can help determine restrictions on the domain that will make the function one-to-one.

Horizontal and Vertical Line Test

Descriptions Examples
The function f(x)=x2f(x)=x^2 is a function, but it does not pass the horizontal line test, so it is not one-to-one. This means the inverse of the function f(x)=x2f(x)=x^2 is not a function.
The graph of the inverse of f(x)=x2f(x)=x^2 is a reflection of f(x)f(x) over the line y=xy=x.

To identify the equation of the inverse, replace f(x)f(x) with yy, and then switch the positions of xx and yy.
y=x2  x=y2y=x^2\ \rightarrow\ x=y^2
The graph of the inverse does not pass the vertical line test, so it is not a function.
If the domain of the function f(x)=x2f(x)=x^2 is restricted to x0x \geq 0, the function passes the horizontal line test, so it is one-to-one.
When the domain of f(x)=x2f(x)=x^2 is restricted to x0x \geq 0, the graph of the inverse passes the vertical line test, so it is a function.

To identify the rule for the function, solve the inverse x=y2x=y^2 for yy.
x=y2±x=y2±x=y\begin{aligned}x&=y^2\\\pm\sqrt{x}&=\sqrt{y^2}\\\pm\sqrt{x}&=y\end{aligned}
Since the domain of the original function was restricted to positive values of xx, the inverse function is only the positive square root, x=y\sqrt{x}=y. Replace yy with f1(x)f^{-1}(x).
f1(x)=xf^{-1}(x)=\sqrt{x}

A vertical line test can help determine whether or not a graph is a graph of a function. A horizontal line test can help identify whether or not the graph of a function is one-to-one. Restricting the domain of a graph of a function ensures that the inverse will be unique.

Step-By-Step Example
Restricting the Domain of a Function
Identify a restriction on the domain of the function so that it has an inverse:
f(x)=(x3)2+2f(x)=\left ( x-3 \right )^{2}+2
Then determine that inverse.
Step 1
Graph the function.
Step 2

Restrict the domain so that the new function is one-to-one.

Limit the xx-values used so that each yy-value appears only once.
So f(x)f(x) is one-to-one for x3x \geq3.
Step 3
Replace f(x)f(x) with yy.
y=(x3)2+2y=\left ( x-3 \right )^{2}+2
Step 4
Switch the positions of xx and yy.
x=(y3)2+2x=\left ( y-3 \right )^{2}+2
Step 5
Solve for yy.
x=(y3)2+2x2=(y3)2+22x2=(y3)2x2=(y3)2x2=y3x2+3=y3+3x2+3=y\begin{aligned}x&=\left ( y-3 \right)^{2}+2\\x-2&=\left ( y-3 \right )^{2}+2-2 \\x-2&=\left ( y-3 \right)^{2}\\\sqrt{x-2}&=\sqrt{\left ( y-3 \right)^{2}}\\\sqrt{x-2}&=y-3\\\sqrt{x-2}+3&=y-3+3\\\sqrt{x-2}+3&=y\end{aligned}
Solution

Replace yy with f1(x)f^{-1}(x), and include the domain.

For x3x \geq3, the inverse is:
f1(x)=x2+3f^{-1}(x)=\sqrt{x-2}+3