Given: f(x) = 2x sin^2x + x cos2x in [0, pi/4]f(x)=2xsin2x+xcos2x∈[0,π4]
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)
