Can someone help me with proving roots?

Question

$F(x) = 2x^3 +x^2−4x−2+cosx$. How do I prove that F(x) has exactly three solutions(roots) and is strictly increasing on $(-\infty,a_1)$ and $(a_2,\infty)$, and strictly decreasing on $(a_1,a_2)$ ? Where $a_1$ and $a_2$ are local maximum or minimum points.

So this is what I have been thinking.

1. I could use IVT I find out that I have $F(x)=0$ in three intervals if i plug in some values for $x$. But still, it would not prove that $F(x)$ have exactly three solutions.
2. If I set $F '(x)=0$ I could prove that there is exactly three solutions and where $F(x)$ is increasing/decreasing, but this equation is not possible to solve.

What should I do?

Answer 1

One long method is that you can start from the third derivative and go back, since $F'''(x)=12 +\sin x$ and you can easily sketch that, now sketch and interpret the variations of $F''$ then $F'$ and finally $F$.

EDIT:

You're right $F'(x)$ has 2 roots that are about $-0.9$ and $0.7$ and between the 2 roots $F'$ is under the $x$-axis, so $F$ is increasing then decreasing (between $-0.9$ and $0.7$) and then increasing, which means that it have a local maximum and a local minimum, and since the maximum is positive and the minimum is negative this means that it have only $3$ distinct roots.

Answer 2

Hint

Consider that $F(x)$ is bounded by $$G(x)= 2x^3 +x^2−4x−1$$ and $$H(x)= 2x^3 +x^2−4x−3$$

Answer 3

use te equation $$\cos(x)=-2x^3-x^2+4x+2$$ and use that $$|\cos(x)|\le 1$$ Ok, you have also $$|-2x^3-x^2+4x+2|\le 1$$