Lesson Objectives
• Demonstrate an understanding of one-to-one functions
• Demonstrate an understanding of how to find the inverse of a function
• Learn how to find the inverse of a domain-restricted function

## How to Find the Inverse of a Domain-Restricted Function

Over the course of the last few lessons, we have learned about one-to-one functions and how to find the inverse of a one-to-one function. We previously learned that a one-to-one function and its inverse will undo each other. Suppose that f(x) and g(x) are inverses. $$f(g(x))=x$$ For every x in the domain of g. $$g(f(x))=x$$ For every x in the domain of f.
In some cases, we need to find an inverse for a function that is not one-to-one. Functions that are not one-to-one functions, known as many-to-one functions, are those that fail the horizontal line test, which means that multiple input values (x-values) can produce the same output value (y-value). In such cases, finding the inverse directly becomes impossible because the inverse function should have a unique output (y-value) for each input value (x-value). Without that uniqueness, we can't reverse the input-output relationship accurately.
To overcome this challenge, we can introduce a domain restriction to make the function one-to-one over a specific interval of input values (x-values). By limiting the domain, we ensure that each input value (x-value) corresponds to a unique output value (y-value), allowing us to find the inverse function. So, the key is to identify a suitable domain restriction that transforms the many-to-one function into a one-to-one function before finding its inverse. Let's begin with a simple example.
Example #1: Find the inverse. $$f(x)=x^2 - 1$$ Looking at the graph above, we can see that our function is not a one-to-one function. We can also see that the vertex occurs at (0, -1). Notice that if we restricted the domain to either x ≥ 0 or x ≤ 0, we would have a one-to-one function with the same range: [-1, ∞).
For simplicity, let's use the domain restriction of x ≥ 0. $$f(x)=x^2 - 1, x ≥ 0$$ Now that we have restricted our domain and created a one-to-one function with the same range as the original, let's follow our procedure for finding the inverse of a function.
1) Replace f(x) with y. $$y=x^2 - 1, x ≥ 0$$ 2) Swap x and y, this means the domain and range will swap. $$x=y^2 - 1, y ≥ 0$$ 3) Solve for y. $$x=y^2 - 1, y ≥ 0$$ Add 1 to each side: $$x + 1=y^2, y ≥ 0$$ Switch sides: $$y^2=x + 1, y ≥ 0$$ Take the square root of each side: $$y=\pm \sqrt{x + 1}, y ≥ 0$$ Do we need the negative square root here? No, we are specifically stating that y is non-negative with our restriction. So we only want the principal square root. $$y=\sqrt{x + 1}$$ At this point, you can drop the restriction.
4) Replace y with f-1(x). $$f^{-1}(x)=\sqrt{x + 1}$$ You can check your work using function composition. $$f(f^{-1}(x))=(\sqrt{x + 1})^2 - 1$$ $$f(f^{-1}(x))=x + 1 - 1$$ $$f(f^{-1}(x))=x$$ $$f^{-1}(f(x))=\sqrt{x^2 - 1 + 1}$$ $$f^{-1}(f(x))=\sqrt{x^2}$$ $$\sqrt{x^2}=|x|$$ $$|x|=\left\{\begin{array}{lr}x, x ≥ 0\\-x, x < 0\end{array}\right.$$ Here, we know that x is non-negative since we restricted the domain of f. $$\sqrt{x^2}=|x|=x, x≥0$$ $$f^{-1}(f(x))=x$$ Additionally, we can create a simple sketch. Recall that a function and its inverse are a reflection across the line y = x. Example #2: Find the inverse. $$f(x)=\frac{1}{x^2}$$ Looking at the graph above, we can see that our function is not a one-to-one function. Notice that if we restricted the domain to either x > 0 or x < 0, we would have a one-to-one function with the same range: (0, ∞). Note, we need a strict inequality here since x can't be zero. This would give division by zero, which is undefined.
To change things up, let's use the domain restriction of x < 0. Note, you can also choose x > 0 (most students find this easier to work with). $$f(x)=\frac{1}{x^2}, x < 0$$ Now that we have restricted our domain and created a one-to-one function with the same range as the original, let's follow our procedure for finding the inverse of a function.
1) Replace f(x) with y. $$y=\frac{1}{x^2}, x < 0$$ 2) Swap x and y, this means the domain and range will swap. $$x=\frac{1}{y^2}, y < 0$$ 3) Solve for y. $$x=\frac{1}{y^2}, y < 0$$ Multiply both sides by y2/x: $$\require{cancel}\frac{y^2}{\cancel{x}}\cdot \cancel{x}=\frac{\cancel{y^2}}{x}\cdot \frac{1}{\cancel{y^2}}, y < 0$$ $$y^2=\frac{1}{x}, y < 0$$ Take the square root of each side: $$y=\pm \sqrt{\frac{1}{x}}, y < 0$$ $$y=\pm \frac{1}{\sqrt{x}}, y < 0$$ Do we need the principal square root here? No, we are specifically stating that y is negative with our restriction. So we only want the negative square root. $$y=- \frac{1}{\sqrt{x}}$$ Let's rationalize the denominator: $$y=- \frac{1}{\sqrt{x}}\cdot \frac{\sqrt{x}}{\sqrt{x}}$$ $$y=- \frac{\sqrt{x}}{x}$$ 4) Replace y with f-1(x). $$f^{-1}(x)=- \frac{\sqrt{x}}{x}$$ You can check your work using function composition. $$f(f^{-1}(x))=\frac{1}{\left(-\frac{1}{\sqrt{x}}\right)^2}$$ $$f(f^{-1}(x))=\large{\frac{\frac{1}{1}}{\frac{1}{x}}}$$ $$f(f^{-1}(x))=\frac{1}{1}\cdot \frac{x}{1}$$ $$f(f^{-1}(x))=x$$ $$f^{-1}(f(x))=-\frac{1}{\sqrt{\frac{1}{x^2}}}$$ $$f^{-1}(f(x))=-\large{\frac{1}{\frac{1}{\sqrt{x^2}}}}$$ $$f^{-1}(f(x))=-\large{\frac{\frac{1}{1}}{\frac{1}{\sqrt{x^2}}}}$$ $$f^{-1}(f(x))=-\frac{1}{1}\cdot \frac{\sqrt{x^2}}{1}$$ $$f^{-1}(f(x))=-\sqrt{x^2}$$ $$f^{-1}(f(x))=-|x|$$ $$|x|=\left\{\begin{array}{lr}x, x ≥ 0\\-x, x < 0\end{array}\right.$$ Because of our domain restriction on our function f, we know that x is negative. $$|x|=-x, x < 0$$ Let's use this to simplify: $$f^{-1}(f(x))=-|x|, x < 0$$ $$f^{-1}(f(x))=-(-x)$$ $$f^{-1}(f(x))=x$$ Additionally, we can create a simple sketch. Recall that a function and its inverse are a reflection across the line y = x. Let's now look at a different type of problem. Suppose we have a one-to-one function, where the domain is restricted due to the nature of the function, such as a square root function. Here, we know that in the real number system, we can't take the square root of a negative. Let's look at an example.
Example #3: Find the inverse. $$f(x)=\sqrt{x - 9}$$ We know this function is a one-to-one function, therefore, it will have an inverse. What happens when we apply our normal procedure?
Step 1) Replace f(x) with y. $$y=\sqrt{x - 9}$$ Step 2) Swap x and y, this means the domain and range will swap. $$x=\sqrt{y - 9}$$ Step 3) Solve for y. $$x=\sqrt{y - 9}$$ Square both sides: $$x^2=(\sqrt{y - 9})^2$$ $$x^2=y - 9$$ Switch sides and add 9 to each side: $$y=x^2 + 9$$ We know that y = x2 + 9 is not a one-to-one function. So what is the actual inverse? We are on the right track, however, we need to include a domain restriction. Let's go back to our original function f(x). $$f(x)=\sqrt{x - 9}$$ $$\text{domain}: [9, \infty)$$ $$\text{range}: [0, \infty)$$ Recall that for an inverse, we swap the domain and range. So the range will be the domain for the inverse. $$f^{-1}(x)=x^2 + 9, x ≥ 0$$ You can check your work using function composition. $$f(f^{-1}(x))=\sqrt{(x^2 + 9) - 9}$$ $$f(f^{-1}(x))=\sqrt{x^2}$$ $$f(f^{-1}(x))=|x|$$ $$|x|=\left\{\begin{array}{lr}x, x ≥ 0\\-x, x < 0\end{array}\right.$$ Because of our domain restriction, we know that x is non-negative. $$|x|=x, x ≥ 0$$ $$f(f^{-1}(x))=x$$ $$f^{-1}(f(x))=(\sqrt{x - 9})^2 + 9$$ $$f^{-1}(f(x))=x - 9 + 9$$ $$f^{-1}(f(x))=x$$ Additionally, we can create a simple sketch. Recall that a function and its inverse are a reflection across the line y = x.

#### Skills Check:

Example #1

Find the inverse. $$f(x)=\sqrt{x - 11}$$

A
No Inverse
B
$$f^{-1}(x)=2x^2 - 11$$
C
$$f^{-1}(x)=x^2 - 11, x ≤ 0$$
D
$$f^{-1}(x)=x^2 + 11, x ≤ 0$$
E
$$f^{-1}(x)=x^2 + 11, x ≥ 0$$

Example #2

Find the inverse. $$f(x)=\frac{1}{\sqrt{x + 2}}$$

A
No Inverse
B
$$f^{-1}(x)=\frac{1}{x^2}- 2, x ≤ 0$$
C
$$f^{-1}(x)=\frac{1}{x^2}- 2, x < 0$$
D
$$f^{-1}(x)=\frac{1}{x^2}- 2, x > 0$$
E
$$f^{-1}(x)=\frac{1}{x^2}- 2, x ≥ 0$$

Example #3

Find the inverse. $$f(x)=x^2 + 4, x ≥ 0$$

Hint: Think carefully about the restrictions before answering.

A
No Inverse
B
$$f^{-1}(x)=\sqrt{x - 4}, x ≥ 0$$
C
$$f^{-1}(x)=-\sqrt{x - 4}, x ≤ 4$$
D
$$f^{-1}(x)=\sqrt{x - 4}, x ≥ 4$$
E
$$f^{-1}(x)=-\sqrt{x - 4}, x ≥ 4$$

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