Question

Show that the equation has exactly one real root.

17. \(2x + \cos x = 0\)

Short answer

The required proof \(2x + \cos x = 0\) is obtained.

Step-by-step solution

Given data

“Let a function\(f\)satisfies the following conditions,

1. A function\(f\)is continuous on the closed interval\((a\;,\;b)\).

2. A function\(f\)is differentiable on the open interval\((a\;,\;b)\).

3.\(f(a) = f(b)\)

Then, there is a number\(c\)in open interval\((a\;,\;b)\)such that\({f^\prime }(c) = 0\).”

Intermediate Value Theorem:

“Suppose that\(f\)is continuous on the closed interval\((a\;,\;b)\)and let\(N\)be any number between\(f(a)\)and\(f(b)\), where\(f(a) \ne f(b)\).

Then, there exists a number\(c\)in\((a\;,\;b)\)such that\(f(c) = N\).”

Calculation to find the interval in which \(x\)

Rearrange the terms of the equation as, \(\cos x = - 2x\).

Solve for \(x\).

Recall the fact that, \(\cos x\) always lies in the closed interval \(( - 1,1)\).

That is, \( - 1 \le \cos x \le 1\).

Since \(\cos x = - 2x\), the above inequality becomes \( - 1 \le - 2x \le 1\).

Dividing the inequality by \( - 2\) yields, \(\frac{{ - 1}}{{ - 2}} \ge \frac{{ - 2x}}{{ - 2}} \ge \frac{1}{{ - 2}}\).

That is, \(\frac{1}{2} \ge x \ge - \frac{1}{2}\).

Thus, it can be concluded that \(x\) lies in the interval \(\left( {\frac{{ - 1}}{2},\frac{1}{2}} \right)\).

Note that, the given function is a combination of polynomial and trigonometric functions.

Since the given function is a polynomial, it is continuous and differentiable.

Calculation to solve the given equation

By Intermediate Value Theorem, there exist at least one root in the interval \(\left( {\frac{{ - 1}}{2},\frac{1}{2}} \right)\) such that \(2x + \cos x = 0\).

Suppose that, if the equation has distinct real roots \(a\) and \(b\), then \(f(a) = f(b) = 0\).

Since \(f\) is continuous on \((a,b)\) and differentiable on \((a,b)\) then by Rolle's Theorem, there is a number \(c\) in ( \(a,b)\) such that \({f^\prime }(c) = 0\).

Identify that \({f^\prime }(x) = 2 - \sin x\).

So, \({f^\prime }(c) = 2 - \sin c\).

Note that \({f^\prime }(c) = 2 - \sin c > 0\) as \(\sin c \le 1\).

Since the derivative of the function is positive, the function is increasing and do not passes the \(x\)-axis twice.

This leads to a contradiction to the assumption that the given equation has distinct real roots.

Therefore, it can be concluded that the given equation has exactly one root.

Hence, the required proof is obtained.