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

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?

• You're not done, you have to show that this $x_0$ is unique. Nov 12, 2019 at 4:03
• Rolle's theorem. Suppose there are two zeros in the interval. Then there would have to be a point in the interval where the derivative was zero. But the derivative is clearly positive in the interval. Nov 12, 2019 at 4:03

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 $$[0,\frac{\pi}{2}]$$.

• That's perfect.
– help
Nov 12, 2019 at 5:35

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.

Hint

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