What are the absolute extrema of #f(x)=2xsin^2x + xcos2x in[0,pi/4]#?

1 Answer
Mar 28, 2018

Answer:

absolute max: # (pi/4, pi/4)#

absolute min: #(0, 0)#

Explanation:

Given: #f(x) = 2x sin^2x + x cos2x in [0, pi/4]#

Find first derivative using the product rule twice.

Product rule: #(uv)' = uv' + v u'#

Let #u = 2x; " "u' = 2#

Let #v = sin^2x = (sin x)^2; " "v' = 2 sin x cos x#

#f'(x) = 2x2 sin x cos x + 2sin^2x + ...#

For the second half of the equation:
Let #u = x; " "u' = 1#

Let #v = cos(2x); " "v' =(-sin(2x))2 = -2sin(2x)#

#f'(x) = 2x2 sin x cos x + 2sin^2x + x(-2sin(2x)) + cos(2x) (1)#

Simplify:
#f'(x) = cancel(2x sin(2x)) + 2sin^2x cancel(-2x sin(2x)) + cos(2x)#

#f'(x) = 2 sin^2x + cos(2x)#

#f'(x) = 2 sin^2x + cos^2x - sin^2x#

#f'(x) = sin^2x + cos^2x #

The Pythagorean Identity #sin^2x + cos^2x = 1#

This means there are no critical values when #f'(x) = 0#

Absolute Maximum and minimums would be found at the endpoints of the function interval.

Test endpoints of the function:

#f(0) = 0; " Absolute minimum:" (0, 0)#

#f(pi/4) = 2*pi/4 sin^2(pi/4) + pi/4 *cos(2*pi/4)#

#f(pi/4) = pi/2 (1/sqrt(2))^2 + pi/4 * cos (pi/2)#

#f(pi/4) = pi/2 * 1/2 + pi/4 * 0#

#f(pi/4) = pi/4; " Absolute maximum:" (pi/4, pi/4)#