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The original function is $f(x)= (3x-2)^.5$

find $y=f^{-1}(x)$ and its domain.

So I found the inverse equation to be $y=((x^2)+2)/3 $ The correct answer for the domain is all reals when $x \geq 0$. Why is this so? Why would it not be all real numbers?

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up vote 3 down vote accepted

The domain of $f^{-1}$ is the range of $f$. Your function is $f(x)=\sqrt{3x-2}$, $x\ge \frac23$. Obviously, $f(x)\ge 0$ and so the range is at most the non-negative reals. Let $y\ge 0$. Then $$y=f(x)\iff y=\sqrt{3x-2}\iff y^2=3x-2\iff x=\frac{y^2+2}3$$ You must check whether or not $x\ge \frac23$. $$x\ge \frac23\iff \frac{y^2+2}3\ge \frac23\iff y^2\ge 0$$ which is true. Thus the domain of $f^{-1}$ (I suppose you have already proven $f$ is 1-1) is $[0,+\infty)$ and $$f^{-1}(x)=\frac{x^2+2}3$$

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+1 for sportsmanship, (Not the badge) – Rustyn Jan 26 '13 at 7:25
@user59714: Note that on $I:[2/3,\infty)$, the function is increasing strictly and so it has an inverse. +1 for straightforward answer. – Babak S. Jan 26 '13 at 7:53

The domain of the inverse function, $f^{-1}(x)$, takes on the range of the function, $f(x)$. Your original function's range is the set of non-negative real numbers, so hence the set of non-negative real numbers must be the domain of $f^{-1}(x)$.

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+1 Nice approach. ;-) – Babak S. Jan 26 '13 at 7:54

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