# Critical numbers of $x^{-5}\log x$

On a homework problem, I got the wrong answer and figured out what to put in for it to be marked correct (online homework), but I am unsure why it is right.

The problem is to find the critical numbers for $f(x)=x^{-5}\log x$. Critical numbers occur where the derivative is 0 or undefined, so the first step is to find the derivative.

Product rule: $-5(x^{-6})\ln(x) + (x^{-5})(1/x)$

Simplify: $-5(x^{-6})\ln(x) + (x^{-6})$

Factor out $(x^{-6})$: $(x^{-6})(-5\ln(x)+1)$

$x^{-6}=1/(x^6)$ which will be undefined at $0$, so that should be part of the list of critical numbers.

Now to find zeros of $-5\ln(x)+1$: $-5\ln(x)+1=0$, $1=5\ln(x)$, $1/5=\ln(x)$, $x=e^{1/5}$.

I put in the list $0,e^{1/5}$ and it was marked incorrect.

On a hunch I removed $0$ from the list. My answer was marked correct.

Isn't $f'(x)$ undefined at $x=0$, or am I missing something? I have been marked correct on other answers which listed points where the derivative is not defined, so I am sure my definition of critical numbers is correct.

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"wolframalpha.com/input/?i=Plot%5Bx%5E%28-5%29*Log%5Bx%5D%5D" (hint) –  Amzoti Nov 14 '12 at 2:13
So f(x) is undefined at 0... I don't understand the significance of this. f'(x) is still undefined at 0. –  Big Endian Nov 14 '12 at 2:18

Critical numbers occurs where the function is defined and the derivative is zero or undefuned. Since the function isn't defined at $x=0$, it can't have a critical point there.