Question

need help with the second derivative of
h(t)=2 + cos(2t)

Answer 1

h'(t) =-2sin(2t)
h"(t)=-4cos(2t)

Answer 2

Where's your 1st derivative :P

Answer 3

Good job :)

Answer 4

-4cost(2t)=0
divide 4 to both sides
cos(2t)=0

Answer 5

h(t)=2 + cos(2t)
h' (t) = - 2* sin 2t
h" (t) = -4 cos 2t

Answer 6

I was thining of using a double angle formula to solve
cos(2t)=cos^2(t)-sin^2(t)
then t=pi/4

Answer 7

Is your question about 2nd derivative?

Answer 8

I think the second derivative tells me if pi/4 is a max or min, not sure

Answer 9

yes, and I was given the interval from - pi to pi

Answer 10

Max/ min determine by 1st derivative!

Answer 11

ok so what does the second derivative tell me??? I am confused.
I thought the first derivative also, told me increasing or decreasing.
Yes I do recall using that information for max and min

Answer 12

ok is it concave up or down or point of inflection?

Answer 13

Sooryyy, precal, think in reverse side, then you'll get it correctly!

Answer 14

Yep, 2nd derivative is for inflex, increase/ decreasing!

Answer 15

then when I solve it, I think it is pi/4 what happens at that point according to the 2nd derivative

Answer 16

well I still not sure what to do

Answer 17

That's the whole point? I still have no clue what your question about? What your purpose of taking second derivative?

Answer 18

If you find min/ max -> let f'(x) = 0

Answer 19

ok I think the question is:
Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. I am helping someone understand the problem. I think they mislead me to do the 2nd derivative.

Answer 20

The purpose of 2nd derivative is find inflex, increase => that what they want

Answer 21

If so let f"(x) = 0