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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?

Mary Star
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    show that $f'(x)$ has only one root on the interval in question – Vasili Mar 16 '18 at 21:37
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    So far so good. So, you just need to show that f'(x) has only one root in the interval. If $f'(x)= 0$ then $2x = 10 \sin x.$ In $(0,\frac {\pi}{2}), 10\sin x > 2x$ and in $(\frac {\pi}{2},\sqrt {10}), 10 \sin x$ is strictly decreasing while $2x$ is strictly increasing, so the two curves can cross at most one time. – Doug M Mar 16 '18 at 21:45
  • Note that as $x^2+10\cos x $ is an even function, you will also prove exactly two negative real roots. – hardmath Mar 16 '18 at 22:31

1 Answers1

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$$f'(x)=2x-10\sin x$$ 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.

Your IDE
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