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

Question

Charles Arocha · Accepted Answer

Start by differentiating.

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

You can use the chain rule on #(sinx)^2#.

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

Critical numbers occur whenever the derivative equals #0#. Hence,

#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 #(0, pi/3)#.

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, #h(x)# is decreasing on #(pi/3, pi)#

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, #h(x)# is increasing on #(pi, (5pi)/3)#

Accordingly, we can deduce that minimums will occur whenever #x = pi +- 2pin# and maximums will occur whenever #x = pi/3 +- 2pin# AND #x = (5pi)/3 +- 2pin#. This is because a maximum is seen at the point where a function stops increasing and begins decreasing, and a minimum is seen at a point where a function stops decreasing and begins increasing.

A graphical verification yields the same result.

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

Hopefully this helps!

Samantha Adkison · Answer

undefined To find the critical numbers of $ h(x) = \sin^2(x) + \cos(x) $, you need to first find the derivative of $ h(x) $ with respect to $ x $, then set it equal to zero and solve for $ x $.

1. Find the derivative of $ h(x) $ using the chain rule and sum rule:

$$ h'(x) = 2\sin(x)\cos(x) - \sin(x) $$

2. Set $ h'(x) $ equal to zero:

$$ 2\sin(x)\cos(x) - \sin(x) = 0 $$

3. Solve for $ x $:

$$ \sin(x)(2\cos(x) - 1) = 0 $$

This equation is true when either $ \sin(x) = 0 $ or $ 2\cos(x) - 1 = 0 $.

Solving $ \sin(x) = 0 $ gives us critical numbers when $ x = 0 $ or $ x = \pi $.

Solving $ 2\cos(x) - 1 = 0 $ gives us critical numbers when $ x = \frac{\pi}{3} $ or $ x = \frac{5\pi}{3} $.

4. Evaluate $ h(x) $ at each critical number to determine whether they correspond to maximum, minimum, or neither.

$$ h(0) = \sin^2(0) + \cos(0) = 0 + 1 = 1 $$
$$ h(\pi) = \sin^2(\pi) + \cos(\pi) = 0 - 1 = -1 $$
$$ h(\frac{\pi}{3}) = \sin^2(\frac{\pi}{3}) + \cos(\frac{\pi}{3}) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} $$
$$ h(\frac{5\pi}{3}) = \sin^2(\frac{5\pi}{3}) + \cos(\frac{5\pi}{3}) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} $$

Thus, the maximum value of $ h(x) $ is $ \frac{5}{4} $ which occurs at $ x = \frac{\pi}{3} $ and $ x = \frac{5\pi}{3} $, and the minimum value of $ h(x) $ is $ -1 $ which occurs at $ x = \pi $.