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The question is: Assume $I = [−1, 3]$ and $f : I → R$ is defined by $f(x) = −x^3 + 3x ^2 + 9x$, $x ∈ I$. Find $J = f(I)$, where $f(I)$ is the range of $f$, that is, $f(I) = \{f(x) : x ∈ I\}$. Explain why $f^{−1}: J → R$ exists and why $f^{−1}$ is strictly increasing and continuous on $J$.

So I already found $J$ which is the min and max of $f$ on the interval. What's throwing me off is proving that $f^{−1}$ is strictly increasing and continuous. My argument is that there is a root for $f$ at $x=0$ which is in $J$, which means that $f$ will go to infinite at $0$. So $f$ would not be continuous, but my teacher assures me it is. And when I try to prove that it is strictly increasing, I find that the derivative is less then $0$. Can anyone shed some light on this?

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    That makes sense now. Thank you2011-11-18

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There is no trouble at $x=0$. We are dealing with the inverse function $f^{-1}$, not the reciprocal $\dfrac{1}{f}$.

Use f'(x) (the derivative) to show that $f$ is increasing in your interval. If you do not wish to use the derivative, you can factor out $x$, and do a more complicated analysis. (The reason you may wish to avoid the derivative is if your course is very rigorous and properties of the derivative have not been derived yet.)