2.5 The Inverse Function: $f^{-1}$

1. Discussion and Formal Definition

The inverse function reverses the action of a function. For example, if $$f(x)=x+10,$$ then $f$ maps $$0\mapsto 10,\qquad 1\mapsto 11,\qquad 2\mapsto 12.$$ The inverse procedure maps the outputs back to the original inputs: $$10\mapsto 0,\qquad 11\mapsto 1,\qquad 12\mapsto 2.$$ Hence the inverse function is $$f^{-1}(x)=x-10.$$

Let $f:\mathbb{R}\to\mathbb{R}$. The inverse function $f^{-1}$ is defined by

$$f(x)=y\quad\Longleftrightarrow\quad f^{-1}(y)=x.$$
  • The domain of $f$ becomes the range of $f^{-1}$, and the range of $f$ becomes the domain of $f^{-1}$: $$D_{f^{-1}}=R_f,\qquad R_{f^{-1}}=D_f.$$
  • The inverse of the inverse function is the original function: $$(f^{-1})^{-1}=f.$$
  • The compositions of inverse functions give the identity function: $$(f^{-1}\circ f)(x)=x,\qquad (f\circ f^{-1})(x)=x.$$
Function Domain Range
$f$ $D_f$ $R_f$
$f^{-1}$ $R_f$ $D_f$

2. Determining the Inverse Function

To find the inverse function algebraically:

  1. Set the function equal to $y$: $f(x)=y$.
  2. Solve algebraically for $x$ in terms of $y$.
  3. Keep the solution, but replace $y$ by $x$.
  4. State the domain of $f^{-1}$ by using the range of $f$.
Step General procedure Example $f(x)=x+10$
1 Set $f(x)=y$ $x+10=y$
2 Solve for $x$ $x=y-10$
3 Replace $y$ by $x$ $f^{-1}(x)=x-10$

EXAMPLE 1

Let $$f(x)=3x+5.$$ Find (a) $f^{-1}(x)$ and (b) $f^{-1}(11)$.

Solution:

(a) Set $f(x)=y$: $$ 3x+5=y \implies 3x=y-5 \implies x=\dfrac{y-5}{3}. $$ Replace $y$ by $x$ and we get $$f^{-1}(x)=\dfrac{x-5}{3}.$$
(b) Substitute $x=11$ into the inverse: $$ f^{-1}(11)=\dfrac{11-5}{3}=\dfrac{6}{3}=2. $$
Alternatively, $f^{-1}(11)$ is the value of $x$ such that $f(x)=11$: $$ f(x)=11 \implies 3x+5=11 \implies 3x=6 \implies x=2. $$ Therefore, $$f^{-1}(11)=2.$$

EXAMPLE 2

Let $$f(x)=2x^2-1,\qquad x\geq 0.$$ Find (a) $f^{-1}(x)$ and (b) $f^{-1}(49)$.

Solution:

(a) Set $f(x)=y$: $$ 2x^2-1=y \implies 2x^2=y+1 \implies x^2=\dfrac{y+1}{2}. $$
Since the original domain is $x\geq 0$, take the positive square root $$x=\sqrt{\dfrac{y+1}{2}}.$$
Replace $y$ by $x$ $$f^{-1}(x)=\sqrt{\dfrac{x+1}{2}}.$$
Since the range of $f(x)=2x^2-1$ for $x\geq 0$ is $y\geq -1$, the domain of $f^{-1}$ is $$x\geq -1.$$
(b) $$ f^{-1}(49)=\sqrt{\dfrac{49+1}{2}}=\sqrt{25}=5. $$ Alternatively, $f^{-1}(49)=x$ means $f(x)=49$, so $$ 2x^2-1=49 \implies 2x^2=50 \implies x^2=25 \implies x=5 $$ where $x=5$ is chosen because the domain is $x\geq 0$.

EXAMPLE 3

Let $$f(x)=\dfrac{x+1}{x+2}.$$ (a) Show that $$f^{-1}(x)=\dfrac{2x-1}{1-x}.$$ (b) Verify that $(f\circ f^{-1})(x)=x$.
(c) Find the domain and range of $f$ and $f^{-1}$.

Solution:

(a) Set $f(x)=y$: $$\begin{aligned} \dfrac{x+1}{x+2}=y &\implies x+1=y(x+2) \implies x+1=yx+2y \\ &\implies x-yx=2y-1 \implies x(1-y)=2y-1 \\ &\implies x=\dfrac{2y-1}{1-y}. \end{aligned}$$ Therefore, $$f^{-1}(x)=\dfrac{2x-1}{1-x}.$$
(b) Verify $(f\circ f^{-1})(x)=x$: $$\begin{aligned} (f\circ f^{-1})(x) &=f\left(\dfrac{2x-1}{1-x}\right)=\dfrac{\dfrac{2x-1}{1-x}+1}{\dfrac{2x-1}{1-x}+2}\\[0.5cm] &=\dfrac{\dfrac{2x-1+1-x}{1-x}}{\dfrac{2x-1+2-2x}{1-x}}=\dfrac{\dfrac{x}{1-x}}{\dfrac{1}{1-x}}=x. \end{aligned}$$
(c) The original function is undefined when $x=-2$, so $$D_f=\mathbb{R}\setminus\{-2\}.$$ Since $f(x)=1$ has no solution, $$R_f=\mathbb{R}\setminus\{1\}.$$
Therefore, $$D_{f^{-1}}=R_f=\mathbb{R}\setminus\{1\},$$ and $$R_{f^{-1}}=D_f=\mathbb{R}\setminus\{-2\}.$$

EXAMPLE 4 (Composite Inverses)

Let $$f(x)=1-2x,\qquad g(x)=\dfrac{1}{x}.$$ Find: (a) $(f\circ g)(x)$
(b) $(g\circ f)(x)$
(c) $(g\circ f^{-1})(x)$
(d) $(f\circ g^{-1})(x)$
(e) $(f\circ g)^{-1}(x)$
(f) $(f^{-1}\circ g^{-1})(x)$

Solution:

(a) $$ (f\circ g)(x)=f(g(x))=f\left(\dfrac{1}{x}\right)=1-\dfrac{2}{x}. $$
(b) $$ (g\circ f)(x)=g(f(x))=g(1-2x)=\dfrac{1}{1-2x}. $$
(c) First find $f^{-1}$: $$ y=1-2x \implies 2x=1-y \implies x=\dfrac{1-y}{2}. $$ Hence, $$f^{-1}(x)=\dfrac{1-x}{2}.$$ Therefore, $$ (g\circ f^{-1})(x) =g\left(\dfrac{1-x}{2}\right) =\left(\dfrac{1-x}{2}\right)^{-1} =\dfrac{2}{1-x}. $$
(d) Since $g(x)=\dfrac{1}{x}$, we have $g^{-1}(x)=\dfrac{1}{x}$. Therefore, $$ (f\circ g^{-1})(x) =f\left(\dfrac{1}{x}\right) =1-\dfrac{2}{x}. $$
(e) Since $$(f\circ g)(x)=1-\dfrac{2}{x},$$ set $$1-\dfrac{2}{x}=y.$$ Then $$ 1-\dfrac{2}{x}=y \implies 1-y=\dfrac{2}{x} \implies x=\dfrac{2}{1-y}. $$ Therefore, $$(f\circ g)^{-1}(x)=\dfrac{2}{1-x}.$$
(f) $$ (f^{-1}\circ g^{-1})(x) =f^{-1}\left(\dfrac{1}{x}\right) =\dfrac{1-\dfrac{1}{x}}{2} =\dfrac{x-1}{2x}. $$
Notice that $$(f\circ g)^{-1}\neq f^{-1}\circ g^{-1}.$$ The correct order is $$(f\circ g)^{-1}=g^{-1}\circ f^{-1}.$$

3. Graph of the Inverse Function

Reflection: The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y=x$.
Coordinate switch: If $(a,b)$ lies on $y=f(x)$, then $(b,a)$ lies on $y=f^{-1}(x)$.
Intersection property: If $f$ is increasing, then $f$ and $f^{-1}$ may intersect only on the line $y=x$. Therefore, to find their intersection points, it is enough to solve $$f(x)=x.$$
Self-inverse functions: A function is called self-inverse if $f^{-1}=f$. Then $$(f\circ f)(x)=x,$$ and its graph is symmetric about $y=x$.

EXAMPLE 5 (Graph of an Inverse)

If $$f(x)=x^2,\qquad x\geq 0,$$ then $$f^{-1}(x)=\sqrt{x}.$$ Their graphs are reflections of each other in the line $y=x$.

Since $f$ is increasing on $x\geq 0$, the intersection points of $f$ and $f^{-1}$ lie on $y=x$. Solve: $$ \begin{aligned} f(x)=x &\implies x^2=x \implies x^2-x=0 \implies x(x-1)=0 \\ &\implies x=0\quad\text{or}\quad x=1. \end{aligned} $$ Therefore, the intersection points are $$(0,0)\qquad\text{and}\qquad (1,1).$$

EXAMPLE 6 (Self-Inverse Function)

The function $$f(x)=\dfrac{1}{x},\qquad x\neq 0,$$ is self-inverse because $$f^{-1}(x)=\dfrac{1}{x}.$$ Hence, $$(f\circ f)(x)=x.$$

4. Presupposition for $f^{-1}$: One-to-One Functions

For an inverse function to exist, the original function must be one-to-one. This means different inputs must give different outputs: $$x_1\neq x_2\quad\Longrightarrow\quad f(x_1)\neq f(x_2).$$

Equivalently, the contrapositive form is often more useful: $$f(x_1)=f(x_2)\quad\Longrightarrow\quad x_1=x_2.$$

Graphically, a function is one-to-one if it passes the horizontal line test: any horizontal line intersects the graph at most once.

EXAMPLE 7 (One-to-One vs Many-to-One)

Compare $f(x)=2x$ and $g(x)=x^2$.

For $f(x)=2x$: $$ f(x_1)=f(x_2) \implies 2x_1=2x_2 \implies x_1=x_2. $$ Therefore, $f(x)=2x$ is one-to-one.
For $g(x)=x^2$, different inputs may give the same output: $$g(2)=4,\qquad g(-2)=4.$$ Therefore, $g(x)=x^2$ is many-to-one on $\mathbb{R}$.
However, if the domain is restricted to $x\geq 0$, then $g(x)=x^2$ becomes one-to-one, and its inverse is $\sqrt{x}$.