AP calculus problem?

Question

i need help with this question. please help me with this.

Assume that f(x)=cos^3x.solve f '(x)=0 for x in the interval [0,2pi]

Answer 1

I'll do a similar one, then you can copy, okay?

f(x) = sin^2(x) , solve f'(x)=0

f' is just the derivative, so by the chain rule,

f'(x) = 2 sin(x) * cos (x)

2 sin(x) * cos(x) = 0 can only happen when any one of the terms is 0. And we are dealing with real numbers in the interval 0 - 2pi.

sin(x) = 0 at x=0, x=pi, x=2pi
cos(x) = 0 at x=pi/2, 3pi/2

Hope that puts you on the path.
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p.s. A valiant effort by the 13-year-old who posted just before me. Her will to try will lead her to do great things, one day.

Answer 2

This uses the general power rule (which is based on the chain rule).

(uⁿ)′= nuⁿ⁻¹u′

In this case, u = cosx and n = 3. So,

f′(x) = −3 cos²x sinx
0 = −3 cos²x sinx

The derivative will have zeroes when cos²x = 0 or sinx = 0. Also, cos²x will have zeroes when cosx = 0. Then we can use our knowledge of the sine and cosine functions:

cosx = 0
x = π/2, x = 3π/2

sinx = 0
x = 0, x = π, x = 2π

So, your five solutions are 0, π/2, π, 3π/2, 2π. (Remember that this is a closed interval from 0 to 2π.)

The below answers by cidyah and ケマン are not correct because their stated derivative is not correct.

Answer 3

i am not sure...but i will try. i just need to know is cos^3x cos to the 3x power or cos to the thrid power times x.

1)the x in the interval[0,2pi] is 0 (i think) so

that would be f(0)=cos^3x so...i guess that would be...yea thats as best as i can do! sorry i only thirteen and i know a little about functions, plus i don't feel like getting my calculator to find the cos :) sorry i couldn't help much

Answer 4

f(x) = cos³ x
f'(x) = -3 sin x cos² x = 0
sin x = 0, x = 0 or x = π or x = 2π (since interval includes 2π)
cos² x = 0
cos x = 0
x = π/2 or x = 3π/2

Answer 5

(a million) For huge style one you will could desire to apply the chain rule and locate the by-manufactured from sin first yet go away what's in the parentheses the way it is: by-manufactured from sin is cos so... cos(x^3-5x^2+10x-8) now multiply by the by-manufactured from what's in the parentheses: (3x^2-10x+10)cos(x^3-5x^2+10x-8) and that could desire to be your answer (2) For huge style 2 comparable situation: first by-manufactured from a capability: 5(x^2sinx)^4 then multiply however the by-product on the interior: *you will could desire to apply the product rule for that* 5(x^2sinx)^4(2xsinx+x^2cosx) and that's your answer

Answer 6

derivative of (cosx)^3 is

f '(x) = -3 sin(x) (cos(x))^2


f'(x) =0 when sin(x) =0 or cos(x) =0

sin(x) =0 ----> x =0; x = pi ; x =2pi
cos(x) =0 ---> x = pi/2 ; x = 3pi/2

Answer 7

f'(x)=-3sin(3x)=0
sin(3x)=0
sin sin(0)=0 sin(pi)=0 and sin(2pi)=0 in [0,2pi],
3x=0 which means x=0
3x=pi which means x=pi/3
3x=2pi which means x=2pi/3
x=0,pi/3 or 2pi/3

Answer 8

Use chain rule
f'(x)=-3sin(3x)cos(3x)
Plug in 0 for all x