Chapter 8.4, Problem 7QR

To determine

To prove: The inequality $\left|\frac{\cos x}{x^2}\right|\le \frac{1}{x^2}$ is true for all $-\infty <x<\infty$ .

Explanation of Solution

Given information: The given inequality is $\left|\frac{\cos x}{x^2}\right|\le \frac{1}{x^2}$ .

Concept used: The range of the cosine function is $\left[-1,1\right]$ .

Proof:

It is known that the modulus function return only positive value of the function. So, the range of the $\left|\cos x\right|$ will be $\left[0,1\right]$ .

The denominators on both sides of the given inequality are same.

Since, the value of $\left|\cos x\right|$ will always be less than or equal to 1, the left side of the inequality which is $\left|\frac{\cos x}{x^2}\right|$ will always be less than or equal to the right side of the inequality which is $\frac{1}{x^2}$ .

Hence Proved.