# What are the number of greatest/least possible maxima for $f(x) = \frac{x^n}{x-1}$?

I have a math contest question that I found in my textbook and I have no idea where to start. Please provide some hints as to how to go about solving this:

Consider the function $f(x) = \frac{x^n}{x-1}$ where $n$ is a positive integer. Let $G$ represent the greatest possible number of local extreme for $f(x)$, and let $L$ represent the least possible number of local extrema. Which statement is true?

a) $G = n$ and $L = n-1$

b) $G = n$ and $L = 1$

c) $G = n-1$ and $L = 0$

d) $G = 2$ and $L = 1$

e) $G = 2$ and $L=0$

Thanks a lot.

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$f(x)$ is smooth, so its extrema can be found by finding where its derivative vanishes. –  Christopher A. Wong Mar 27 '13 at 23:25

Simply set the derivative equal to zero and count the unique roots. By the quotient rule, $$f'(x)=\frac{nx^{n-1}(x-1)-x^{n}}{(x-1)^{2}}=0$$ Note $x \ne 1$; count $x=0$ as a solution and divide by $x^{n-1}$. Hence, $f'(x)=0$ is equivalent to
$$n(x-1)+x=0 \implies (n-1)x=n \implies x=\frac{n}{n-1}$$ $x=0$ is always a stationary point, and $x=\frac{n}{n-1}$ is if and only if $n \ne 1$.
If $n$ is odd, then $x=0$ is an inflection point - otherwise it is a local maximum.
The case $n=1$ shows the answer immediately (perhaps try to sketch it?).
the derivative of $f$ is zero at $0$ and $\frac{n}{n-1}$ so these are the possible local extrema. At $\frac{n}{n-1}$ the derivative goes from negative to positive, so it's a minimum. The behavior at zero depends on whether $n$ is even or odd.