# Continuity and strictly increasing on an interval

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?

-
$f^{-1}$ means the inverse of $f$, not the reciprocal of $f$. – André Nicolas Nov 18 '11 at 1:18
Your argument that $f$ goes infinite at $0$ is based on confusing $f^{-1}$ with $1/f$. $f^{-1}$ is the inverse function, not the reciprocal: it’s the function with the property that for every $x\in I$, $f^{-1}(f(x))=x$, and for every $x\in J$, $f(f^{-1}(x))=x$. Use $f\;'$ to show that $f$ is strictly increasing on $I$ and think about why this means that $f^{-1}$ must be strictly increasing on $J$. – Brian M. Scott Nov 18 '11 at 1:31
That makes sense now. Thank you – rioneye Nov 18 '11 at 1:42

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.)