What are the critical values, if any, of # f(x)= x^2sinx +sinxcos^2x in [0,pi]#?

1 Answer
Jul 11, 2018

Answer:

#x ~~2.3339#

Explanation:

Given: #f(x) = x^2sin x + sin x cos^2 x " from "[0, pi]#

One way to find the critical values is to graph and find the maximum using a graphing calculator :

relative max: #(2.333931, 4.2818209)#
graph{x^2sin x + sinx( cos x)^2 [-2, 3.14159, -2, 5]}

The second way is to find the first derivative and set it equal to zero and solve for #x#:

Find the first derivative using the product rule: #(uv)' = uv' + v u'#

For the first part of the function let #u = x^2; " "u' = 2x, " "v = sinx; " "v' = cos x#

#d/(dx)(x^2sin x) = x^2 cosx + 2x sin x#

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

#d/(dx)(sin x cos^2 x) = cos x (-2cos x sin x) + (cos x)^2cos x#

# = -2cos^2 x sin x + cos^3 x#

#f'(x) = x^2 cosx + 2x sin x -2cos^2 x sin x + cos^3 x#

Find critical values : #f' = 0#

#f'(x) = x^2 cosx + 2x sin x -2cos^2 x sin x + cos^3 x = 0#

This is a difficult problem. The easiest way to solve is to use a graphing calculator to graph the derivative and then solve for the zero (#x#-intercept).

graph{x^2 cos x + 2x sin x -2(cos x)^2 sin x + (cos x)^3 [-2, 3.14159, -2, 5]}

#"zero at " x ~~2.3339#