I got the function $$f(x)=\begin{cases}{x^2+2,\text{if }x\in(-2, 0]}\\{3x-7,\text{if }x\in(0, 3)}\end{cases}$$ and I want to find the $f^{-1}(x)$. How to do that? I find the inverse of the $2$ parts separate or somethin else?

  • $\begingroup$ Find the image of the function, and check if it is injective. If it is, the inverse of the function takes the image back to the original domain injectively. $\endgroup$ – George Sep 20 '17 at 14:17
  • $\begingroup$ You just compute the inverses of each part, then combine them accordingly. You can verify that it is indeed the inverse by showing $f\circ f^{-1}=Id=f^{-1}\circ f$. Both parts of the function are injective on their interval of definition. You just need to make sure the codomain is the same as the image of $f$ to ensure surjectivity. $\endgroup$ – Dave Sep 20 '17 at 14:18

For the first part. Write $y=x^2+2$. Then $y\in [2,6)$. Now swap $x$ and $y$. Then you get $x=y^2+2$ so $y=\pm \sqrt{x-2}$ and $x\in [2,6)$. Choose part with $-$ because $x<0$ (in OP). Do the second part your self.

$$f^{-1}(x)=\begin{cases}{-\sqrt{x-2},\text{if }x\in[2,6)}\\{{x+7\over 3},\text{if }x\in(-7, 2)}\end{cases}$$


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