How do you find all local maximum and minimum points using the second derivative test given #y=cos^2x-sin^2x#?

Question

Elijah Abram · Accepted Answer

#y(x)# has a local maximum for #x=kπ# and a local minimum for #x=kπ+π/2#

As:

#y=cos^2x-sin^2x=cos(2x)#

we do not really need derivatives because we know that #cos(2x)# has a local maximum for #x=kpi# where #cos(2x)=1# and a local minimum for #x=kpi+pi/2# where #cos(2x)=-1#.

However we can check:

#y'(x) = -2sin(2x)#

so the critical points occur for:

#2x= kpi# or #x=kpi/2#

#y''(x)= -4cos(2x)#

and in fact:

#y''(2kpi) = -4 < 0# so these points are maximums, and

#y''((2k+1pi)) = 4 > 0# so these points are minimums.

Christopher Agresta · Answer

undefined To find the local maximum and minimum points using the second derivative test for $ y = \cos^2(x) - \sin^2(x) $:

1. Find the first derivative, $ y' $.
$$ y' = \frac{d}{dx}(\cos^2(x) - \sin^2(x)) = -2\cos(x)\sin(x) $$

2. Find the critical points by setting the first derivative equal to zero and solving for $ x $.
$$ -2\cos(x)\sin(x) = 0 $$
$$ \Rightarrow \cos(x) = 0 \text{ or } \sin(x) = 0 $$

3. Solve for $ x $ for each condition:
$$ \cos(x) = 0 \Rightarrow x = \frac{\pi}{2} + n\pi, \text{ where } n \in \mathbb{Z} $$
$$ \sin(x) = 0 \Rightarrow x = n\pi, \text{ where } n \in \mathbb{Z} $$

4. Calculate the second derivative, $ y'' $.
$$ y'' = \frac{d}{dx}(-2\cos(x)\sin(x)) = -2(\sin^2(x) - \cos^2(x)) $$

5. Evaluate $ y'' $ at each critical point:
$$ \text{For } x = \frac{\pi}{2} + n\pi, \text{ where } n \in \mathbb{Z}: $$
$$ y'' = -2(\sin^2(\frac{\pi}{2} + n\pi) - \cos^2(\frac{\pi}{2} + n\pi)) $$
$$ \text{For } x = n\pi, \text{ where } n \in \mathbb{Z}: $$
$$ y'' = -2(\sin^2(n\pi) - \cos^2(n\pi)) $$

6. Determine the nature of the critical points using the second derivative test:
- If $ y'' > 0 $, the critical point is a local minimum.
- If $ y'' < 0 $, the critical point is a local maximum.
- If $ y'' = 0 $, the test is inconclusive.

7. Plug in the critical points into the second derivative and determine the nature of each critical point.