Show that $\cos x=x^3+x^2+4x$ has exactly one root in $[0,\frac{\pi}{2}]$

Question

This is the section of the textbook on derivatives. I'm not sure if I should be using Rolle's Theorem or if Mean Value Theorem would be useful here. Intermediate Value Theorem?

I understand we can write this as $x^3+x^2+4x-\cos x=0$. We know this is uniformly continuous since it's just the sum of continuous functions on a compact domain. All the other benefits of working with a uniformly continuous function follow. I'm just not sure how to do this.

Maybe, we can write this as $f(x)=g(x)$ where $f(x), g(x)$ are differentiable and uniformly continuous on their domain. We can write $h(x)=g(x)-f(x)$ for all $x\in [0,\frac{\pi}{2}]$. I think we evaluate the endpoints of the interval, so $h(0)=-1$ and $h(\frac{\pi}{2})=\frac{\pi^3+2\pi^2+16\pi}{8}$. There's clearly a sign change, so by the IVT, somewhere in the interval, $h(x_0)=0$. Is that the end of the proof or do I have to show this $x_0$ is unique, how do I show it's exactly one?

Answer 1

Let $f(x)=-\cos(x)+x^3+x^2+4x$. Note $f(0)=-1<0$ and $f(\frac{\pi}{2})>0$. Therefore, by the Intermediate Value Theorem $f$ must admit a zero in $[0,\frac{\pi}{2}]$. If $f$ had another zero in $[0,\frac{\pi}{2}]$ then from Rolle's theorem $f'(x)=\sin(x)+3x^2+2x+4$ would admit a zero between those two zeros of $f$ but $f'$ is strictly positive in $[0,\frac{\pi}{2}]$. Therefore, $f$ only admits one zero on $[0,\frac{\pi}{2}]$.

Answer 2

Let $f_1(x) = \cos(x)$ and $f_2(x)=x^3+x^2+4x$. Note that, over $[0, \pi/2]$.

$$f_1’(x) =-\sin x < 0$$

$$f_2’(x) = 3x^2+2x +4 > 0$$

That is $f_1(x)$ strictly decreases and $f_2(x)$ strictly increases. Given that $f_1(0)=1$, $f_1(\pi/2)=0$ and $f_2(0) =0$, the two curves have to cross and cross once within $(0,\pi/2)$, hence having only one root.

Answer 3

Hint

$\cos x$ is concave and strictly decreasing and $x^3+x^2+4x$ is convex and strictly increasing, both over $\left(0,{\pi\over 2}\right)$