I got
$$f^{-1}(x) = (x-2)^2$$
Is this answer right?
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3yes it is right – Jul 01 '14 at 17:46
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1Take notice of the fact that it is true if $f:\left[0,\infty\right)\rightarrow\left[2,\infty\right)$ and $f^{-1}:\left[2,\infty\right)\rightarrow\left[0,\infty\right)$. The domain of $f$ must be the codomain of $f^{-1}$ and the codomain of $f$ must be the domain of $f^{-1}$. Domain and codomain are essential parts of the function, which is (too) often neglected. – drhab Jul 01 '14 at 18:00
3 Answers
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Yes, your answer is correct. For future users:
\begin{align} y &= \sqrt{x} + 2 \\ x &= \sqrt{y} + 2 \\ (x-2) &= \sqrt{y} \\ (x-2)^2 &= y \end{align}
The idea is to replace $x$ with $y$ and solve for $y$.
Jeel Shah
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Assuming you do mean $\sqrt{x}+2$ and not $\sqrt{x+2}$, the answer is correct, except that I'd add that the domain of $f^{-1}$ is $x\ge 2$.
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For $x \geq 0$ you have $f(x) =y \geq 2$ and $$y= \sqrt{x}+2 \iff \sqrt{x}=y-2 \iff x=(y-2)^2=:f^{-1}(y).$$
Surb
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