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I have to find maximum and minimum of the function $x^n\ln^k x$ in the interval $\bigl[\frac{1}{e},e\bigr]$ with $k$ and $n \in N$

and I'm lost in the calculus.

$y'= x^{n-1} \ln^{k-1}x (n\ln x+k)$

$y'=0 \Leftrightarrow x=1 \lor x= e^{-\tfrac{k}{n}}$

While $1 \in\bigl[\frac{1}{e},e\bigr]$, for $e^{-\tfrac{k}{n}}$ it depends.

If $k<-n$ then $x=e^{-\tfrac{k}{n}}<e$

If $k>n$ then $x=e^{-\tfrac{k}{n}}<\dfrac{1}{e}$

If $-n<k<n$ then $\dfrac{1}{e}<x=e^{-\tfrac{k}{n}}<e$

From this moment I'm lost and I can't decide where $y'>0$ Can someone help me?

Bernard
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Anne
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1 Answers1

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I think you solved the problem already because $k \ge 0$ and $n > 0$

To be the second optimum $e^{-\tfrac{k}{n}}$ must satisfy that $e^{-1} \le e^{-\tfrac{k}{n}}\le e$ which means $-n \le k \le n$ or in other words $0 < n \le k$

I would just ignore the other cases where $e^{-\tfrac{k}{n}}$ is not in the interval $[e^{-1},e]$

What could be still missing from your point of view?

Whether maximum or minimum in the case that you have 2 optima depends on the comparison between $e^{-\tfrac{k}{n}}$ and $1$.

Why did you want to investigate $y'>0$ if you already have the zeros of it?

Noureddine Ouertani
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  • yes, indeed. I compared the values and found a solution, also if I would not ignore the case of only one point with y'=0 – Anne Nov 01 '20 at 20:59