Question

I am looking for critical points and I can't remember how to x in the function f '(x) = -5sin(2x)

Answer 1

solve f'=0

Answer 2

wait you want to find the critical numbers of f not f' right?

Answer 3

To find critical numbers of f
you find f'
and then solve f'=0 and find when f' dne
any of those x values you find within the domain of f is a critical number for f

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To find the critical numbers of f'
you find f''
and then solve f''=0 and find when f'' dne
any of those x values you find within the domain of f' is a critical number for f'

Answer 4

f(x)=5cos^2x, I already found f '(x) i was just stuck at that point

Answer 5

ok well set f'=0 and solve for x

Answer 6

that is -5sin(2x)=0
so sin(2x)=0 is what you need to solve

Answer 7

do you know when sin(u)=0

Answer 8

no, how do you calculate that?

Answer 9

sin(u)=0
use unit circle

Answer 10

sin(0)=0
sin(pi)=0
sin(2pi)=0
and so on...

Answer 11

so if u=npi
then sin(2x)=0 gives a zeros when 2x=n pi
solve for x