On the interval (0, 2 $\pi$), f(x) = | sin x cos(9x)| has
A) a relative maximum only
B) a relative minimum only
C) both a relative maximum and a relative minimum
D) no relative extrema
E) cannot be determined

Question

Quizplus · Accepted Answer

To find the relative extrema of f(x), we have to find all critical points of f(x) on the interval (0, 2π/10).

The critical points occur where either sin x = 0 or cos(9x) = 0.

When sin x = 0, we have x = kπ for some integer k. However, this does not produce any critical points since f(x) is not differentiable at these points (the absolute value function is not differentiable at 0).

When cos(9x) = 0, we have x = π/18 + kπ/9 for some integer k. Since f(x) is periodic with period 2π/9, we only need to consider the values of x in the interval (0, 2π/9).

At x = π/18, f(x) has a relative minimum since sin x > 0 and cos(9x) < 0.

At x = 5π/18, f(x) has a relative maximum since sin x < 0 and cos(9x) < 0.

Therefore, f(x) has both a relative maximum and a relative minimum on the interval (0, 2π/10), so the answer is C.

On the Interval (0, 2 π\piπ ), F(x) = | Sin X Cos(9x)| Has A) a Sin

On the Interval (0, 2 $\pi$ ), F(x) = | Sin X Cos(9x)| Has
A) a Sin