# Maximum value of $f(x)=x^3-3x$ in $\mathbf{R}$ satifying $x^4+36 \le 13x^2$

Problem:

Find the maximum value of $f(x) = x^3-3x$ on the set of all real numbers $x$ satisfying $x^4+36 \le 13x^2$.

I made a graph of $f(x)$ (but I don't know how to show it here).

I know that the solution set of the inequality is $-3\le x \le -2$ and $2 \le x \le 3$, but after that I am a bit lost.

Do I just plug in the value of those solution sets of the inequality to find which one is largest for $f(x)$?

-

Since $x$ must satisfy $x^4 + 36 \leq 13x^2$, we get the solution set $-3 \leq x \leq -2$ and $2 \leq x \leq 3$, or $x \in [-3,-2]\cup[2,3]$.

If $f(x)$ is continuous on a closed and bounded interval, then by the extreme value theorem the maximum of $f$ exists. It is either a local maximum in the interior of the domain, or it lies on the boundary of the domain.

I suggest that you find $f'(x)$, see if $f'(x) = 0$ for any $x$ in your domain. If not, the maximum lies on the boundary (i.e. $x = -2, x = -3,x = 2$ or $x = 3$).

Hope that helps!

-

Differentiating gives us $f'(x)=3x^2-3$.

The domain: $x^4+36≤13x^2$ iff $0\le -x^4+13x^2-36$ iff $0\le -(x^2)^2+13x^2-30$ iff $0\le -(x-3)(x+2)(x-2)(x+3)$, so I'm getting that your domain is $[-3,2]\cup [2,3]$

Therefore, you want the solutions to $f'(x)=0$ on this domain, if they exists. $f'(x)=0$ has solutions $x=-1, x=1$, so now we need to know which one, if either of them, gives rise to a maximum.

-