# How do you find the critical numbers for #h(x) = sin^2 x + cos x# to determine the maximum and minimum?

##### 1 Answer

Start by differentiating.

#h(x) = (sinx)^2 + cosx#

You can use the chain rule on

#h'(x) = 2sinxcosx - sinx#

Critical numbers occur whenever the derivative equals

#0 = 2sinxcosx - sinx#

#0 = sinx(2cosx - 1)#

If we solve, we get

#sinx = 0 or cosx = 1/2#

This means that

#x = 0, pi, pi/3, (5pi)/3#

Now let's select test points in between to determine where the function is increasing/decreasing.

**Test point 1: #x = pi/6#**

#h'(pi/6) = sin(pi/3) - sin(pi/6)#

#h'(pi/6) = sqrt(3)/2 - 1/2#

#h'(pi/6) = (sqrt(3) - 1)/2#

This is positive, so the function is increasing on

**Test point 2: #x = pi/2#**

If we evaluate within the derivative, we get:

#h'(pi/2) = sin(2(pi/2)) - sin(pi/2)#

#h'(pi/2) = sin(pi) - sin(pi/2)#

#h'(pi/2) = 0 - 1#

#h'(pi/2) = -1#

Hence,

**Test point 3: #x = (3pi)/2#**

#h'((3pi)/2) = sin(2(3pi)/2) - sin((3pi)/2)#

#h'((3pi)/2) = sin(3pi) - sin((3pi)/2)#

#h'((3pi)/2) = 0 - (-1)#

#h'((3pi)/2) = 1#

Hence,

Accordingly, we can deduce that minimums will occur whenever

A graphical verification yields the same result.

graph{y = (sinx)(sinx) + cosx [-22.8, 22.83, -11.4, 11.38]}

Hopefully this helps!