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i have encountered myself with a mathematic question problem. The exercise says: The function $f$ is defined by $f:xa \sqrt{3-2x}$. Evaluate $f^{-1}(5)$. Does anyone have a clue how to resolve the equation. What does "a" represent? Does it involve the function?

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Maybe it should be (or was) $f \colon x \mapsto \sqrt{2-2x}$. And the symbol $\mapsto$ or similar thing was confused in transcription. –  GEdgar Sep 25 '11 at 19:32
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I can't tell what $a$ is supposed to be. It looks like you have a definition of $f(x)=\sqrt{3-2x}$ (note: please do not mix $x$ and $X$). Then $f^{-1}(5)$ is the value of $x$ such that $f(x)=5$, so you would need to solve $5=\sqrt{3-2x}$. Can you do that?

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Let's suppose that definition of function is $f(x)=\sqrt{3-2x}$ than $f^{-1}(x)$ should be inverse function of $f(x)$ at $x=5$ so we need to find $f^{-1}(x)$

$x=\sqrt{3-2f^{-1}(x)} \Rightarrow x^2= 3-2f^{-1}(x) \Rightarrow f^{-1}(x)=\frac{3-x^2}{2} \Rightarrow f^{-1}(5)=\frac{3-5^2}{2} \Rightarrow f^{-1}(5)=-11$

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