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Chapter 1: Problem 70

Determine whether the function has an inverse function. If it does, then find the inverse function. $$f(x)=\sqrt{x-2}$$

Short Answer

Expert verified
Yes, the function \(f(x)=\sqrt{x-2}\) has an inverse function and the inverse function is \(f^{-1}(x)=x^2 + 2\) for \(x \geq 0\).

Step by step solution

01

Write the Function with 'y'

Write the function \(f(x)=\sqrt{x-2}\) as \(y=\sqrt{x-2}\).
02

Interchange Roles of 'x' and 'y'

Next, interchange the roles of 'x' and 'y' to get \(x=\sqrt{y-2}\).
03

Solve for 'y'

To solve for 'y', first square both sides to get rid of the square root. This yields \(x^2 = y-2\). Then, add 2 to both sides to isolate 'y': \(y = x^2 + 2\).
04

Check if the Assumed Inverse is the Inverse

Either \(f(f^{-1}(x)) =x \) or \(f^{-1}(f(x)) =x \) everywhere in their domain. Without loss of generality, let's consider the first one. Substituting \(x^2+2\) into our original function yields \(f(f^{-1}(x)) =\sqrt{(x^2+2)-2}=\sqrt{x^2}=|x|\). We can see \(|x|\neq x\) for \(x<0\), which is in the domain of \(f^{-1}\), so the assumed inverse is not a valid inverse. However, if we restrict the domain of \(f^{-1}(x)\) to \(x \geq 0\) so that \(|x|\equiv x\), then they are inverses. Hence, the original function does have an inverse, which is \(y=x^2 + 2\) for \(x \geq 0\).

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