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.
$$a)\hspace{.2em}f(x)=\sqrt{x + 5}, x ≥ -5$$
$$b)\hspace{.2em}f(x)=\sqrt{2x - 3}, x ≥ \frac{3}{2}$$
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#2:
Instructions: find the inverse of each.
$$a)\hspace{.2em}f(x)=\frac{1}{\sqrt{x + 9}}, x > -9$$
$$b)\hspace{.2em}f(x)=\frac{1}{\sqrt{x - 5}}, x > 5$$
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#3:
Instructions: find the inverse of each.
$$a)\hspace{.2em}f(x)=x^2 + 3, x ≥ 0$$
$$b)\hspace{.2em}f(x)=x^2 - 4x - 1, x ≤ 2$$ Hint: Think about how you can use the quadratic formula.
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#4:
Instructions: find the inverse of each.
$$a)\hspace{.2em}f(x)=x^4 + 2, x ≤ 0$$
$$b)\hspace{.2em}f(x)=(x - 1)^4 - 1, x ≥ 1$$
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#5:
Instructions: find the inverse of each.
$$a)\hspace{.2em}f(x)=\frac{x^2 + 3}{3x^2}, x > 0$$
$$b)\hspace{.2em}f(x)=\frac{2x^2 + 1}{-x^2}, x < 0$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}f^{-1}(x)=x^2 - 5, x ≥ 0$$ 
$$b)\hspace{.2em}f^{-1}(x)=\frac{x^2 + 3}{2}, x ≥ 0$$ 
Watch the Step by Step Video Solution
#2:
Solutions:
$$a)\hspace{.2em}f^{-1}(x)=\frac{1}{x^2}- 9, x > 0$$ 
$$b)\hspace{.2em}f^{-1}(x)=\frac{1}{x^2}+ 5, x > 0$$ 
Watch the Step by Step Video Solution
#3:
Solutions:
$$a)\hspace{.2em}f^{-1}(x)=\sqrt{x - 3}$$ 
$$b)\hspace{.2em}f^{-1}(x)=-\sqrt{x + 5}+ 2$$ 
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#4:
Solutions:
$$a)\hspace{.2em}f^{-1}(x)=-\sqrt[4]{x - 2}$$ 
$$b)\hspace{.2em}f^{-1}(x)=\sqrt[4]{x + 1}+ 1$$ 
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#5:
Solutions:
$$a)\hspace{.2em}f^{-1}(x)=\frac{\sqrt{3(3x - 1)}}{3x - 1}$$ 
$$b)\hspace{.2em}f^{-1}(x)=\frac{\sqrt{-x - 2}}{x + 2}$$ 
