Question

find the absolute maxima, minima of sin(2x)-2cos(x) on {-pi/2 , pi/2}

Answer 1

\[\sin(2x)-2\cos(x)\]

Answer 2

i took the derivative

Answer 3

\[\cos(2x)+\sin(x)=0 is what i get\]

Answer 4

someone in my class solved it by doing this: cos(2x)=sin(x) ...

Answer 5

why and how?

Answer 6

You can do with derivative
expand cos2x = cos^2 -sin^2 +sinx
1-2sin^ 2+sin = 0
(sinx-1)(2sinx +1) = 0
sinx = 1/2 or sinx =1

Answer 7

sorry sinx = -1/2 and sinx =1

Answer 8

so you used a trig identity?

Answer 9

yes cos2x = cosx^2 -sinx^2
and sin^ +cos^ =1

Answer 10

ok, i see what you did. you found 2 critical points right?

Answer 11

Yes one for max value and other for min

Answer 12

alright, and are there other ways to solve this. like using: cos(2x)=sin(x)

Answer 13

umm I am not sure from where you got cos(2x)=sin(x)

Answer 14

that's where i'm confused, this guy solved it somehow like this

Answer 15

and i can barely read his writing

Answer 16

from our equation we will get cos(2x) = -sin(x)

Answer 17

exactly..

Answer 18

are you sure his method gave him right solution.

Answer 19

\[Let f(x)=\sin 2x-2\cos x\]
\[f \prime \left( x \right)=2\cos 2x+2\sin x \]
f'(x)=0 gives
\[2\cos 2x+2\sin x=0, or \cos 2x+\sin x=0\]
\[1-2\sin ^{2}x+\sin x=0\]
\[2\sin ^{2}x-\sin x-1=0\]
\[2\sin ^{2}x-2\sin x+\sin x-1=0\]
\[2 \sin x\left( \sin x-1 \right)+1\left( \sin x+1 \right)=0\]
\[\left( \sin x+1 \right)\left( 2\sin x+1 \right)=0\]
\[\because given interval is \left( \frac{ -\pi }{ 2 },\frac{ \pi }{ 2 } \right)\]
\[either sinx=-1=-\sin \frac{ \pi }{ 2 }=\sin \left( \frac{ -\pi }{ 2 } \right),x=\frac{ -\pi }{ 2 }\]
\[\sin x=\frac{ -1 }{ 2 }=-\sin \frac{ \pi }{ 6 }=\sin \frac{ -\pi }{ 6 },x=\frac{ -\pi }{6 }\]
\[substitute x=\frac{ -\pi }{6 },\frac{ -\pi }{2 },\frac{ \pi }{ 2 }\]
in f(x) and find the absolute max. & min. value

Answer 20

hm, alright, thanks. i'll just do it this way

Answer 21

this is the correct way.

Answer 22

@surjithayer (sinx+1)(2sinx+1)=0 this part of your solution is wrong

Answer 23

there should be a - 1 instead of +1

Answer 24

ok so, how do i find the absolute maxima and minima
i plugged in -pi/2 and pi/2 in my original function and get 0...what do i do

Answer 25

i'm guessing the minima is -pi/2 and maxima is pi/2 in the original function

Answer 26

nice

Answer 27

but i was told to find the absolute extrema, not local

Answer 28

i'm so confused, maybe i should take break

Answer 29

@rowa tell me what's the problem?

Answer 30

i'm trying to graph it, i don't know how to know if it's going up or down.
and if i have a domain of the derived function equal to 0 that is the same domain as the endpoint of my closed interval, can i use it..blablabla