Question

For the function f(x) = \(\frac{sin(x)}{x^2}\) How many points exist in the interval [0, 7π] Such that f'(c) = 0.

(a) 8

(b) 0

(c) 7

(d) 6

The question was asked in class test.

The question is from Rolle’s Theorem in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (d) 6

Best explanation: We know that sine is a periodic function and it is divided by x^2.

Observe that the sine takes the value of zero at integral arguments of π hence at every interval of the form [nπ, (n + 1)π]

We have f(nπ) = f((n + 1) π)

The sine and the polynomial combination is continuous and differentiable at every point except x = 0

Every such interval has a point such that f'(c) = 0

Hence, by Rolles theorem, in every interval of the form [nπ, (n + 1)π] we must have a point such that f'(c) = 0

Leaving the interval [0, π] we are left with six such intervals from 0 to 7π.