Last question for today I promise. I have the function $f(x) = x^3\ln(x) - x$. I have to find its absolute or local max and min, if any. So I studied the derivative but it's not a function I can properly solve, when imposing it equals to zero. So I thought about proving things in different analytical way. The fact is that I am stuck in certan parts.
So furst of all I have
$$f'(x) = 3x^2\ln(x) + x^2 - 1$$
which is continuous in its domain (the same of $f(x)$). I can see "by hand" that $x = 1$ is a solution of $f'(x) = 0$. The fact is that I want indeed to prove that $x = 1$ is a solution, that is there exist a solution. Then I want to prove it's unique. And then I would like to show that that solution is indeed $x = 1$.
So I thought about considering $g(x) = f'(x)$ and hence studying in terms of $g(x)$, that is:
$g'(x) = 5x + 6x\ln(x)$, observing that $g'(x) > 0$ for $x > e^{-5/6}$. This means $g(x)$ is increasing for $x = e^{-5/6}$ and consequently $f(x)$ is convex in that same interval. Since there is a change in the monotonicity of $g(x) = f'(x)$, then $f(x)$ has a minimum (due to how the monotonicity changes) and only one, due to having shown there is only one solution for $g'(x)$, since the solution $x = 0$ is trashed because of Fermat's theorem).
Then I showed there exists only one minimum.
How can I show that this minima is indeed at $x = 1$ without just saying that "it's obvious"?
You can use the second derivative test $f”(1)=5$ is clearly positive and hence the critical point at $x=1$ is a local minimum. BTW there is a much easier way to show that there is only one critical point think about when $3x^2\ln(x)+x^2$ is greater than $1$ and when it is less than $1$.