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I need to find the $x$-value of the extrema point of $$x^2(a\ln x -1)$$

in terms of $a$.

However, the only thing they tell me about $a$ is that $a ≠ 0$. I found the $x$-value of the point, however my difficulty is finding whether the point is maxima or minima. If I would know if $a$ is less than $0$ or greater than $0$ then it would help a lot. However, this is not the case...

$$f'(x) = 2x(a\ln x -1) + x^2\frac{a}{x} = 2x(a\ln x -1) + ax $$ $$f'(x) = 0$$

$$x = e^\frac{2-a}{2a}$$

$$f''(x) = 2a\ln x+3a-2$$ $$f''\left(e^\frac{2-a}{2a}\right) = 2a\cdot\frac{2-a}{2a}+3a-2 = 2-a+3a-2 = 2a$$

I get $2a$. Therefore, it does seem to depend on the value of $a$. (If it's $ a < 0 $ or $ a > 0$).

There is only one answer!

Can someone help me? Thank you very much!

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    Why not discuss in two cases: $a>0$ and $a<0$? It seems that you know how to go on. – 2017-01-08
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    @Jack That's what I thought but there is only one answer, not two. – 2017-01-08
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    What did you find for $f'(x)=0$? Maybe you can share your work? "but there is only one answer, not two" What is the answer then? – 2017-01-08
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    @Jack I updated mi question. – 2017-01-08
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    The answer is what you got there: $\;x=e^{\frac{2-a}{2a}}\;$ , but it is a loocal maxima point if $\;a<0\;$ , or a local minima one if $\;a>0\;$ . – 2017-01-08
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    @DonAntonio as I said there is only one answer – 2017-01-08
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    @PichiWuana Yes, I know: the extrema is $\;x=e^{\frac{2-a}{2a}}\;$ , but in some cases it is a minima and in some others a maxima...or you mean that there's only one case: maxima or minima? Which one? – 2017-01-08

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Your work is good, you're just misinterpreting the meaning of the second derivative.

The derivative is $$ f'(x)=2x(a\ln x-1)+x^2\frac{a}{x}= x(2a\ln x-2+a) $$ which only vanishes for $$ \ln x=\frac{2-a}{2a}=\frac{1}{a}-\frac{1}{2}=g(a) $$ so for $x=e^{g(a)}$. With the second derivative: $$ f''(x)=2a\ln x-2+a+2a=2a\ln x-2+3a $$ so $$ f''(e^{g(a)})=2a\left(\frac{1}{a}-\frac{1}{2}\right)-2+3a=2a $$ which is positive (absolute minimum) for $a>0$ and negative (absolute maximum) for $a<0$.

Computing $f(e^{g(a)})$: $$ f(e^{g(a)})=e^{2g(a)}\left(a\frac{2-a}{2a}-1\right)=-\frac{a}{2}e^{2g(a)} $$ confirms the results, because $$ \lim_{x\to0}f(x)=0 $$ and $$ \lim_{x\to\infty}f(x)= \begin{cases} \infty & \text{if $a>0$}\\ -\infty & \text{if $a<0$} \end{cases} $$