Consider the mapping diagram of the function f(x), shown below.

Let's now reverse the arrows, as follows:

Now, we have a mapping diagram that "reverses" the action of the function f. The diagram itself represents a new function, which we call the inverse of the function f, and denote by f^{-1}.

We can evaluate the inverse function f^{-1} at any point in the range of f. For example, to find the value of f^{-1}(2), we notice that we have an arrow from {\color{black}2} (on the right) to {\color{black}0} (on the left). Therefore, f^{-1}({\color{black}2}) = {\color{black}0}.

In the original function, f, that same arrow went from {\color{black}0} (on the left) to {\color{black}2} (on the right). So, we have that f({\color{black}0})={\color{black}2}, and as a result, we obtain f^{-1}(f(0)) = 0. So, applying the function f and then "reversing" the action using f^{-1} will get us back to 0.

In general, this is true for every x in the domain of the function f\mathbin{:} f^{-1}(f(x)) = x

FLAG

The function f(x) is represented by the mapping diagram above. What is f^{-1}(-1)?

EXPLANATION

Method 1

The function f^{-1}(x) reverses the action of f(x), as depicted below.

From the above, we see that f^{-1}(-1) = 4.

Method 2

From the given mapping diagram, we have

f(4) = -1.

Applying f^{-1} to both sides of the above equation, and using the fact that f^{-1}\left(f(x)\right) = x, we get

\begin{align*} f^{-1}\left(f(4)\right) &= f^{-1}(-1)\\[3pt] 4 &=f^{-1}(-1). \end{align*}

Therefore, we conclude that f^{-1}(-1) = 4.

FLAG

The function $f(x)$ is represented by the mapping diagram above. What is $f^{-1}(4)?$

a
$-\dfrac 1 2$
b
$\dfrac 1 4$
c
$2$
d
$-4$
e
$5$

The function $f(x)$ is represented by the mapping diagram above. What is $f^{-1}(3)?$

a
$-\dfrac 1 3$
b
$-3$
c
$2$
d
$-1$
e
$\dfrac 1 3$
x -8 -4 -1 1 2 3 5
f(x) 7 3 0 -2 -3 -4 -6

Given the function f(x) defined by the table above, what is the value of f^{-1}(3)?

EXPLANATION

From the table, we have

f(-4) = 3.

Applying f^{-1} to both sides of the above equation, and using the fact that f^{-1}\left(f(x)\right) = x, we get

\begin{align*} f^{-1}\left(f(-4)\right) &= f^{-1}(3)\\[3pt] -4 &=f^{-1}(3). \end{align*}

Therefore, we conclude that f^{-1}(3) = -4.

FLAG
$x$ $-9$ $-6$ $-3$ $0$ $3$ $6$ $9$
$f(x)$ $9$ $3$ $2$ $0$ $-2$ $-9$ $-20$

\[ \]

Given the function $f(x)$ defined by the table above, what is the value of $f^{-1}(-9)?$

a
$6$
b
$9$
c
$-2$
d
$20$
e
$-9$

Given that \[ f(0) = 3, \quad f(3) = 6, \quad f(6) = 0, \]

find the value of $f^{-1}(3).$

a
$6$
b
$-6$
c
$-3$
d
$0$
e
$-1$

The graph of y = f(x) is given above. The points A,B,C, and D lie on the graph, as shown. What is the value of f^{-1}(-3)?

EXPLANATION

The point D on the graph has coordinates (2,-3). Therefore,

f(2) = -3.

Applying f^{-1} to both sides of the above equation, and using the fact that f^{-1}\left(f(x)\right) = x, we get

\begin{align*} f^{-1}\left(f(2)\right) &= f^{-1}(-3)\\[3pt] 2 &=f^{-1}(-3). \end{align*}

Therefore, we conclude that f^{-1}(-3) = 2.

FLAG

The graph of $y = f(x)$ is given above. The points $A, B, C, D,$ and $E$ lie on the graph, as shown. What is the value of $f^{-1}(1)?$

a
$2$
b
$-2$
c
$-\dfrac 1 2$
d
$0$
e
$-1$

The graph of $y = f(x)$ is given above. The points $A,B,C,$ and $D$ lie on the graph, as shown. What is the value of $f^{-1}(1)?$

a
$3$
b
$-1$
c
$-\dfrac 1 3$
d
$2$
e
$\dfrac 1 3$

Recall that the f^{-1} "reverses" the action of f so that f^{-1}(f(x)) = x, which can be visualized using the mapping diagram, as shown below.



Since f^{-1}(f(x)) is just the composition (f^{-1} \circ f)(x), we can now give the algebraic definition of the inverse.

The function f^{-1} is called the inverse of the function f if (f^{-1} \circ f)(x) = x for each x in the domain of f.

In other words, to show that two functions f(x) and g(x) are inverses of each other, we need to check the following condition: (g\circ f)(x) = x.

FLAG

Suppose that f(x)=\dfrac{x}{4}-2 and g(x)=4x-2k. If g is the inverse of f, then what is the value of k ?

EXPLANATION

If g is the inverse of f, then we must have

g(f(x)) = x.

To find g(f(x)), we write down the function g and replace every occurrence of x with f(x)\mathbin{:}

\begin{align} g(x) &= 4{\color{blue}x}-2k\\[3pt] g(f(x)) &= 4{\color{blue}f(x)} - 2k \end{align}

Then, we substitute {\color{blue}f(x)} = {\color{blue}\dfrac{x}{4}-2} into the right-hand side of our expression and simplify, as follows:

\begin{align} g(f(x)) &= 4{\color{blue}f(x)} - 2k\\[5pt] &=4\left({\color{blue}\dfrac{x}{4}-2}\right)-2k\\[5pt] &=x - 8 - 2k \end{align}

Because g is the inverse of f, we also know that g(f(x)) = x. Equating the above expression with x and solving for k, we get

\begin{align*} \require{cancel} x - 8 - 2k &= x\\[3pt] \cancel{x} - 8 - 2k &= \cancel{x}\\[3pt] -8-2k &=0\\[3pt] 2k&=-8 \\[3pt] k &=-4. \end{align*}

FLAG

Suppose that $f(x)=x-6$ and $g(x)=x + k.$ If $g$ is the inverse of $f,$ then $k=$

a
$-\dfrac{1}{6}$
b
$\dfrac{1}{6}$
c
$-6$
d
$1$
e
$6$

Suppose that $f(x)=2x$ and $g(x)=\dfrac x k.$ If $g$ is the inverse of $f,$ then $k=$

a
$-2$
b
$\dfrac{1}{2}$
c
$2$
d
$1$
e
$-\dfrac{1}{2}$

Suppose that $f(x)=\dfrac{x}{3}+7$ and $g(x)=3x+k.$ If $g$ is the inverse of $f,$ then $k=$

a
$-7$
b
$-21$
c
$21$
d
$7$
e
$3$
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