I want to prove that the equation $x^2+10\cos x=0$ has exactly two positive roots. I also want to calculate approximations for each root using the bisection method.
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I have done the following:
$f(x)=0 \Rightarrow x^2+10\cos x=0 \Rightarrow x^2=-10\cos x\leq 10 \Rightarrow x\leq \sqrt{10}$
That means that the positive roots will be in the intervall $[0, \sqrt{10}]$, right?
We have that $f(0)=10>0$ and $f(\pi )=\pi^2-10<0$. So, from Bolzano's Theorem we have that the function has a root in $[0,\pi]$.
We have that $f(\pi )=\pi^2-10<0$ and $f(\sqrt{10} )=10+10\cos (\sqrt{10})>0$. So, from Bolzano's Theorem we have that the function has a root in $[\pi, \sqrt{10}]$.
We suppose that there is also a third root in $[0, \sqrt{10}]$. Let these three roots be $x_1<x_2<x_3$. Then we have that $f(x_1)=f(x_2)=f(x_3)=0$.
From Rolle's theorem we have that there are $a\in [x_1, x_2]$ and $b\in [x_2, x_3]$ such that $f'(a)=f'(b)=0$.
Is so far everything correct?
Gow could we get a contradiction?
$$f'(x)=2x-10\sin x$$ It is a bit difficult to determine where exactly the sign changes for the derivative, but it can be observed that it changes somewhere between $ \frac{\pi}{12}$ to $\frac{\pi}{6}$ from positive to negative.
Also, proceeding further, the sign again changes from negative to positive somewhere between $\frac{3 \pi}{4} $ and $\frac{5\pi}{6}$ and maintains the same sign till $\sqrt{10}$. SO, there exists only one root for $f'(x)$ means there are exactly two positive roots for $f(x)$
P.S.: It is interesting to note that it is an even function and there are two more roots on the negative x-axis which are additive inverse of these positive roots.