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Show that the equation below has exactly one root:

$$2x+\cos x=0$$

How would I find the root?

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    Don't you mean IVT, not MVT? – 2017-01-16
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    @S.C.B. well this is under the MVT chapter – 2017-01-16
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    Is the IVT in this chapter chapter also? – 2017-01-16
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    yes my bad it is – 2017-01-16

5 Answers 5

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If $f(x) = 2x+\cos x$ has two zeros, $a$ and $b$, then we have, by MVT (or Rolle's Theorem)

$$0=\frac{f(b)-f(a)}{b-a} = f'(c)$$

for some $c$ between $a$ and $b$. However, $f'(x) = 2-\sin x$ is always positive, so no such $c$ can exist.

To see that there is at least one root, we just plug in $x=10$ and $x=-10$ and apply IVT.

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    thanks, can you tell how i would apply IVT – 2017-01-16
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    Since $f(10)$ is at least $19$ and $f(-10)$ is less than $-19$, and since $f$ is continuous, IVT guarantees a root between $-10$ and $10.$ – 2017-01-16
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Suppose $2x+\cos x$ had [at least] two roots. Then the mean value theorem implies that the derivative is zero somewhere between those two roots. However, the derivative is $2-\sin x$, which is strictly positive. So this shows the number of roots is either $0$ or $1$. But this doesn't completely answer the question...

There is a more direct way. Since the derivative is strictly positive, $2x+\cos x$ is strictly increasing. Using the IVT will give you one root.

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Let $f(x) = 2x + \cos(x)$. Then $f'(x) = 2 - \sin(x) \gt 0$ for all $x$, since $|\sin(x)| \le 1$. This means it crosses the $x$ axis once and never looks back.

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If you derive the function you get:

$$2-\sin x>0\ \forall x \in \mathbb R$$

So the function is strictly increasing. If you take the limit of $2x+\cos x$ to $\pm\infty$ you get $\pm\infty$, hence being the function continuous and strictly increasing it must intersect the $x-axis$ only in a point.

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Previous answers and comments explained that the root is unique.

Now, the question is : what is the zero of $$f(x)=2x+\cos(x)$$ Such equation which mixes polynomila and trigonometric terms do not show analytical solutions and numerical methods should be used. The simplest is probably Newton method which, starting from a guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In the present case, this will write $$x_{n+1}=\frac{x_n \sin (x_n)+\cos (x_n)}{\sin (x_n)-2}$$ Being lazy, let us choose $x_0=0$; then the method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & -0.50000000000000000000 \\ 2 & -0.45062669307724304657 \\ 3 & -0.45018364757777474250 \\ 4 & -0.45018361129487381641 \\ 5 & -0.45018361129487357304 \end{array} \right)$$