### About Inverse of a Domain Restricted Function:

In some cases, we need to find the inverse of a function when the domain is restricted. This will come up when trying to find the inverse of a function with a square root, or if you are asked to find the inverse of a quadratic function where the domain is restricted to create a one-to-one function.

Test Objectives

- Demonstrate an understanding of how to find the inverse of a function
- Demonstrate an understanding of how to find the inverse of a domain restricted function

#1:

Instructions: find the inverse of each.

$$a)\hspace{.2em}f(x)=\sqrt{x + 5}, x ≥ -5$$

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#2:

Instructions: find the inverse of each.

$$a)\hspace{.2em}f(x)=\sqrt{2x - 3}, x ≥ \frac{3}{2}$$

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#3:

Instructions: find the inverse of each.

$$a)\hspace{.2em}f(x)=\frac{1}{\sqrt{x + 9}}, x > -9$$

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#4:

Instructions: find the inverse of each.

$$a)\hspace{.2em}f(x)=x^2 + 3, x ≥ 0$$

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#5:

Instructions: find the inverse of each.

$$a)\hspace{.2em}f(x)=x^2 - 4x - 1, x ≤ 2$$

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Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=x^2 - 5, x ≥ 0$$

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#2:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\frac{x^2 + 3}{2}, x ≥ 0$$

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#3:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\frac{1}{x^2}- 9, x > 0$$

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#4:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=\sqrt{x - 3}, x ≥ 3$$

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#5:

Solutions:

$$a)\hspace{.2em}f^{-1}(x)=-\sqrt{x + 5}+ 2, x ≥ - 5$$