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))#
