Question

Find the local extrema of f on [0, 2pi] and the subintervals on which f is increasing or is decreasing.

f(x) = 2cosx + sin 2x

Answer 1

f' = -2sinx+2cos2x = 0
=>sinx = cos2x => pi/2 - x = 2n(pi) + 2x
=>-pi(4n-1)/6 = x

Answer 2

note:the set ive given might be a subset of the actual solution set.since i've forgotten the cos equivalence conditions

Answer 3

\[f'(x)=-2sinx+2\cos(2x)\]
\[f'=0\]
\[-2sinx+2\cos(2x)=0\]
-----------------------------------------
Think:
remember cos(2x)=cos(x+x)=cosx*cosx-sinx*sinx=cos^2x-sin^2x
also remember sin^2x+cos^2x=1 so cos^2x=1-sin^2x
so we can write cos(2x)=1-sin^2x-sin^2x=1-2sin^2x
-----------------------------------------
so we have
\[-2sinx+2(1-2\sin^2x)=0\]
\[-2sinx+2-4\sin^2x=0\]
\[-4\sin^2x-2sinx+2=0\]
if you want it might be easy for you to write sin^2x as (sinx)^2
\[-4(sinx)^2-2sinx+2=0\]
let u=sinx so u^2=sin^2x
\[-4u^2-2u+2=0\]
\[2u^2+u+1=0\]
so we need to find u
and then we can find x

Answer 4

oops i made an error
it is
\[2u^2+u-1=0\]

Answer 5

\[u=\frac{-1 \pm \sqrt{1-4(2)(-1)}}{2(2)}=\frac{-1 \pm \sqrt{1+8}}{4}=\frac{-1 \pm \sqrt{9}}{4}=\frac{-1 \pm 3}{4}\]

Answer 6

\[u=\frac{-1+3}{4}=\frac{1}{2}, u=\frac{-1-3}{4}=-1\]

Answer 7

remember u=sinx
\[sinx=\frac{1}{2}, sinx=-1\]

Answer 8

first equation gives
\[x=\frac{\pi}{6}+2n \pi,\frac{5 \pi}{6}+2n \pi \]
second equation gives
\[x=\frac{3\pi}{2}+2n \pi\]
for n=...,-2,-1,0,1,2,...

Answer 9

but since we are only looking at 0 to 2pi
we have x=pi/6,5pi/6, 3pi/2

Answer 10

so there are numbers where we have horizontal tangents

Answer 11

|--------|-------|------|---------|
0 pi/6 5pi/6 3pi/2 2pi
plug in numbers between each critical number and endpoint
(except before the first endpoint and after the last endpoint)

Answer 12

any questions?

Answer 13

f'(.5)=+ (we can actually just plug in 0 instead)
f'(pi/2)=-
f'(pi)=+
f'(11pi/6)=+
so increasing on intervals: (0,pi/6)U(5pi/6,3pi/2)

Answer 14

so at pi/6 we had it went from increasing before to decreasing after so we have a max occuring at pi/6

we have a min occuring at 5pi/6 since it goes from decreasing to increasing

Answer 15

oh yeah and it is decreasing on the interval (pi/6,5pi/6)