What are the local extrema and inflection points for #y = 2sin (x) - cos^2 (x)# for #[0, 2π]#?

Question

Emily Ammerman · Accepted Answer

Let #f(x)=y#  # y = f(x) = 2sin (x) - cos^2 (x)# for #[0, 2pi]# To find local extrema:
Find critical numbers for #f#
#y' = 2cosx + 2cosx sinx# #y'# never fail to exist and 
#y'=0# when 
#2cosx + 2cosx sinx = 2cosx(1+sinx)=0# In #[0, 2pi]# this happens at #pi/2# and at #(3pi)/2# Testing:
#y' =  2cosx(1+sinx)=0#, and the factors: #2# and #1+sinx# are always non-negative, so the sign of #y'# matches the sign of #cosx# On #[0, pi/2)# , #y'# is positive (because cosine is) On #(pi/2, (3pi)/2)# , #y'# is negative (because cosine is) On #((3pi)/2, 2pi)# , #y'# is positive (because cosine is) So #f(pi/2) = 2# is a local maximum and #f((3pi)/2) = -2# is a local minimum. To find inflection points:
Investigate the sign of #y''#
#y'' = -2sin (x) + (-2sin^2 (x) +2cos^2x)#
#y'' = -2sin (x) + (-2sin^2 (x) +2(1-sin^2x)#
#y'' = 2-2sinx-4sin^2x#
#y'' = -2(2sin^2x +sinx-1) = -2(2sinx-1)(sinx+1)# #y'' = 0# when #-2(2sinx-1)(sinx+1) = 0# And that happens at  #x= pi/6, (5pi)/6, (3 pi)/2# The factors of #y''#:  #-2# is always negative and #sinx+1# is never negative, so the sign of #y''# will be the opposite of the sign of #2sinx-1# On #[0, pi/6)# , #y''# is positive (#2sinx-1# is negative) On #(pi/6, (5pi)/6)# , #y''# is positive (#2sinx-1# is negative) On #((5pi)/6, (3pi)/2)# , #y''# is positive (#2sinx-1# is negative) On #((3pi)/2, 2pi)# , #y''# is positive (#2sinx-1# is negative) There are inflection points because the concavity changes: #(pi/6 , f(pi/6))#  and #((5pi)/6 , f((5pi)/6))#

Evelyn Agard · Answer

undefined To find the local extrema and inflection points of $ y = 2\sin(x) - \cos^2(x) $ on the interval $[0, 2\pi]$, we first find the critical points by setting the derivative equal to zero and solving for $x$. Then we test these critical points to determine whether they are local extrema or inflection points by analyzing the second derivative.

1. Find the first derivative:
$$ \frac{dy}{dx} = 2\cos(x) + 2\sin(x)\cos(x) $$

2. Set the first derivative equal to zero and solve for $x$:
$$ 2\cos(x) + 2\sin(x)\cos(x) = 0 $$
$$ \cos(x)(2 + 2\sin(x)) = 0 $$
$$ \cos(x) = 0 $$ or $ 2 + 2\sin(x) = 0 $

For $ \cos(x) = 0 $, solutions are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

For $2 + 2\sin(x) = 0$, we get $ \sin(x) = -1 $, which gives $x = \frac{3\pi}{2}$.

So, the critical points are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

3. Find the second derivative:
$$ \frac{d^2y}{dx^2} = -2\sin(x) + 2\cos^2(x) - 2\sin^2(x) $$

4. Evaluate the second derivative at the critical points:
$$ \frac{d^2y}{dx^2} \left(\frac{\pi}{2}\right) = -2 $$
$$ \frac{d^2y}{dx^2} \left(\frac{3\pi}{2}\right) = -2 $$

Since the second derivative is negative at both critical points, they correspond to local maxima.

There are no inflection points on the interval $[0, 2\pi]$.

Therefore, the local extrema for $y = 2\sin(x) - \cos^2(x)$ on $[0, 2\pi]$ are:
- Local maximum at $x = \frac{\pi}{2}$ with $y = 2$
- Local maximum at $x = \frac{3\pi}{2}$ with $y = 2$