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)
The concavity changes, so there are inflection points:
(pi/6 , f(pi/6)) and ((5pi)/6 , f((5pi)/6))
