On the interval (0, 2 $\pi$), f(x) = | sin x cos(2x)| 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 need to find where its derivative is equal to zero or undefined. Thus, we need to find f'(x). 
Using the chain rule and the product rule, we have: 
f'(x) = cos(x)cos(2x) - sin(x)sin(2x) = cos(x)(cos(2x) - 2sin(x)sin(x)) = cos(x)(cos(2x) - sin^2(x))

We can see that f'(x) = 0 at x = pi/4 and x = 3pi/4. 
At x = pi/4, 
f''(x) = -2cos(x)sin(2x) + cos(2x)sin(x) = -sin(4x) - sin(2x) = -sin(4pi/4) - sin(2pi/4) = -1 < 0. 
Thus, f(x) has a relative maximum at x = pi/4.

At x = 3pi/4, 
f''(x) = sin(4x) - sin(2x) = sin(4pi/4) - sin(2pi/4) = 1 > 0. 
Thus, f(x) has a relative minimum at x = 3pi/4.

Therefore, f(x) has both a relative maximum and a relative minimum on the given interval.

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

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