<s>[INST] Find all the intervals in which $ -x^4 + x + 3 \ge 0 $ How do I find all the intervals in which $$-x^4 + x + 3 \ge 0$$?
First of all, I let $f(x) = -x^4 + x + 3$. Then I used the derivative $f'(x) = -4x^3 + 1$ to study its growth.
I used the expression $a^3-b^3 = (a-b)(a^2+ab+b^2)$ to write $f'(x)$ as the following
$$f'(x) = (1 - x\sqrt[3]{4})(1 + x \sqrt[3]{4} + (x \sqrt[3]{4})^2 )$$
from which I clearly have the solution $x = 1/\sqrt[3]{4}$.
For $4^\frac{2}{3}x^2  + x\sqrt[3]{4} + 1$  we have $\Delta = 4^\frac{2}{3}-4\cdot4^\frac{2}{3}\cdot1 < 0 $ which means that there are no solutions in $\mathbb{R}$.
As $f'(2) < 0$ it means that the function is decreasing on $(1/\sqrt[3]{4}, \infty)$ and increasing on $(-\infty, 1/\sqrt[3]{4})$.
But was thinking about an approach to use Rolle's theorem but to no avail as I won't be able to pinpoint solutions of the equation.
 [/INST] You can find the roots of the derivative in a simpler way:
$$-4x^3+1=0\implies x^3=\frac14$$ and in the reals the cubic root is unique.
Now if we evaluate the polynomial at $\dfrac1{\sqrt[3]4}$ we find
$$-\dfrac1{4\sqrt[3]4}+\dfrac1{\sqrt[3]4}+3,$$ which is positive.
So there is a single interval where the polynomial is positive, defined by the two roots, which are on either side of  $\dfrac1{\sqrt[3]4}$. By trial an error, we see that this interval contains $-1$ and $1$ but not $-2$ nor $2$.
The roots do have a closed-form expression, but it is a little masochistic.
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