Inverse Functions: Learn It 3

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Inverse Functions: Learn It 3

Graphing Inverse Functions

Let’s consider the relationship between the graph of a function $f$ and the graph of its inverse.

Consider the graph of $f$ shown in Figure 9(a) and a point $(a,b)$ on the graph.

An image of two graphs. The first graph is of “y = f(x)”, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs “y = f(x)” with the point (a, b), but also graphs the function “y = f inverse (x)”, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”. — Figure 9. (a) The graph of this function $f$ shows point $(a,b)$ on the graph of $f$ . (b) Since $(a,b)$ is on the graph of $f$ , the point $(b,a)$ is on the graph of $f^{-1}$ . The graph of $f^{-1}$ is a reflection of the graph of $f$ about the line $y=x$ .

Since $b=f(a)$ , then $f^{-1}(b)=a$ . Therefore, when we graph $f^{-1}$ , the point $(b,a)$ is on the graph. As a result, the graph of $f^{-1}$ is a reflection of the graph of $f$ about the line $y=x$ .

How to: Graph the Inverse of a Function

Plot the Function: Graph the original function $f(x)$ and plot a few key points.
Reflect Over Line: Reflect these points over the line $y=x$ to find the corresponding points on $f^{-1}$ .
Draw the Inverse: Connect these reflected points to graph the inverse function.
Check: Ensure that each point $(a,b)$ on the original function corresponds to the point $(b,a)$ on the inverse function.
Line of Symmetry: The line $y=x$ should act as a line of symmetry between the function and its inverse.

For the graph of $f$ in the following image, sketch a graph of $f^{-1}$ by sketching the line $y=x$ and using symmetry. Identify the domain and range of $f^{-1}$ .

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). — Figure 10. Graph of $f(x)$ .

Reflect the graph about the line $y=x$ . The domain of $f^{-1}$ is $[0,\infty)$ . The range of $f^{-1}$ is $[-2,\infty)$ . By using the preceding strategy for finding inverse functions, we can verify that the inverse function is $f^{-1}(x)=x^2-2$ , as shown in the graph.

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from -2 to 2. The graph is of two functions. The first function is “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). The second function is “f inverse (x) = (x squared) -2”, an increasing curved function that starts at the point (0, -2). The x intercept is at the approximate point (1.4, 0) and the y intercept is at the point (0, -2). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”. — Figure 11. Graph of $f(x)$ and its inverse.

Restricting Domains

As we have seen, $f(x)=x^2$ does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of $f$ such that the function is one-to-one. This subset is called a restricted domain.

By restricting the domain of $f$ , we can define a new function $g$ such that the domain of $g$ is the restricted domain of $f$ and $g(x)=f(x)$ for all $x$ in the domain of $g$ . Then we can define an inverse function for $g$ on that domain.

For example, since $f(x)=x^2$ is one-to-one on the interval $[0,\infty)$ , we can define a new function $g$ such that the domain of $g$ is $[0,\infty)$ and $g(x)=x^2$ for all $x$ in its domain. Since $g$ is a one-to-one function, it has an inverse function, given by the formula $g^{-1}(x)=\sqrt{x}$ .

On the other hand, the function $f(x)=x^2$ is also one-to-one on the domain $(−\infty,0]$ . Therefore, we could also define a new function $h$ such that the domain of $h$ is $(−\infty,0]$ and $h(x)=x^2$ for all $x$ in the domain of $h$ . Then $h$ is a one-to-one function and must also have an inverse. Its inverse is given by the formula $h^{-1}(x)=−\sqrt{x}$ (Figure 13).

An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is “g(x) = x squared”, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is “g inverse (x) = square root of x”, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is “h(x) = x squared”, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is “h inverse (x) = -(square root of x)”, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”. — Figure 13. (a) For $g(x)=x^2$ restricted to $[0,\infty), \, g^{-1}(x)=\sqrt{x}$ . (b) For $h(x)=x^2$ restricted to $(−\infty,0], \, h^{-1}(x)=−\sqrt{x}$ .

restricted domain

Some functions don’t have inverses over their full domains because they’re not one-to-one. By restricting the domain, we ensure the function is one-to-one. Once the domain is restricted, we can define an inverse.

Consider the function $f(x)=(x+1)^2$ .

Sketch the graph of $f$ and use the horizontal line test to show that $f$ is not one-to-one.
Show that $f$ is one-to-one on the restricted domain $[-1,\infty)$ . Determine the domain and range for the inverse of $f$ on this restricted domain and find a formula for $f^{-1}$ .

The graph of $f$ is the graph of $y=x^2$ shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, $f$ is not one-to-one.

Figure 14. Graph of the function $f(x)=(x+1)^2$ .
On the interval $[-1,\infty), \, f$ is one-to-one.

Figure 15. Looking at a restricted domain graph of a function.

The domain and range of $f^{-1}$ are given by the range and domain of $f$ , respectively. Therefore, the domain of $f^{-1}$ is $[0,\infty)$ and the range of $f^{-1}$ is $[-1,\infty)$ .

To find a formula for $f^{-1}$ , solve the equation $y=(x+1)^2$ for $x$ . If $y=(x+1)^2$ , then $x=-1 \pm \sqrt{y}$ . Since we are restricting the domain to the interval where $x \ge -1$ , we need $\pm \sqrt{y} \ge 0$ . Therefore, $x=-1+\sqrt{y}$ . Interchanging $x$ and $y$ , we write $y=-1+\sqrt{x}$ and conclude that $f^{-1}(x)=-1+\sqrt{x}$ .

Watch the following video to see the worked solution to this example.

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