Question

For f(x) = sin^2 x - cos x and 0≤x≤π, find the value(s) of x for which....

1. f(x) has a local maximum. Write EXACT answer. If none, write NA.
x=_______

2. f(x) has a local minimum. Write EXACT answers in ascending order.
x=_______, x=_______

3. f(x) has a global maximum. If there are none, write NA.
Write EXACT answer.
x=_______

4. f(x) has a global minimum. If there are none, write NA.
Write EXACT answer.
x=________

so find derivative first? then set equal to 0? :/

thanks!!!

Answer 1

not quite sure how to find the derivative of this one :/

Answer 2

use chain rule

Answer 3

\(\large f(x) = \sin^2 x - \cos x \)

Answer 4

u familiar wid chain rule right ?

Answer 5

yes:)

so would it be

sinx * 2cosx + 1 ?

Answer 6

not sure if i did that right though :/

Answer 7

yup ! whats wid +1 in the end ?

Answer 8

Oh, you tried to factor out sinx is it ?

Answer 9

you're right :)

Answer 10

yay!! :D

Answer 11

what will be the next step ?

Answer 12

umm set equal to 0? :/

Answer 13

yes!

Answer 14

\(\large f(x) = \sin^2 x - \cos x \)


\(\large f'(x) = 2\sin x(\sin x)' + \sin x \)

\(\large \qquad= 2\sin x \cos x + \sin x \)

\(\large \qquad= \sin x (2\cos x + 1) \)

Answer 15

corrected the typo^

Answer 16

so we get this? \[sinx(2cosx+1) =0\]

and you get sinx =0 and 2cos x + 1 =0 ?

Answer 17

Looks like you have 3 points to solve for. Go for it!

Answer 18

Yes ! keep going :)

Answer 19

three points? :/

and so x=0 for sinx =0

and cos x = 1/2... x= 1/2cos ?

Answer 20

sin(x) = height of point on unit circle

Answer 21

\(\large \sin x = 0\)

when, \(x = 0\) and \(x = \pi\), right ?

Answer 22

yes:)

Answer 23

so that gives u two solutions

work out the other factor : \(\large 2\cos x + 1 = 0 \)

\(\large \cos x = -\frac{1}{2}\)
\(\large x = ?\)

Answer 24

\[-\frac{ 1 }{ 2\cos }\] ?

Answer 25

lol no

Answer 26

\(\large \cos x\) is a SINGLE FUNCTION. you cannot break it like that.

Answer 27

for example :

if
\(\large f(x) = \frac{1}{2}\)

then,

do u say : \(\large x = \frac{1}{2f}\) ?

Answer 28

ohh okay ahh i see

so would it be 2* cos?

Answer 29

\(\large \cos x = -\frac{1}{2} \)


\(\large x = \cos^{-1}(-\frac{1}{2})\)

Answer 30

\(\large \cos^{-1}\) is the inverse function of \(\large \cos\)

Answer 31

ohh okay:)

Answer 32

that gives \(\large x = \frac{2\pi}{3}\)