Question

Determine the concavity of the graph of f(x)=3sin(x) + 4(cos(x))^2

Answer 1

Find f''(x) first.
Find the domains where f''(x) is positive. f(x) will be concave up in that domain.
Find the domains where f''(x) is negative. f(x) will be concave down in that domain.

Answer 2

Okay, but deriving it is confusing for me, for f'(x) I got, 3cos(x) + 8cos(x)sin(x), is that right?

Answer 3

It should be 3cos(x) + 8cos(x)(-sin(x)) = 3cos(x) - 8cos(x)sin(x)

Answer 4

Right, sorry I missed that. Okay for F"(x) is it -3sin(x)-8cos^2(x)-8sin^2(x)?

Answer 5

I mean +8sin^2(x)

Answer 6

correct. Write cos^2(x) as 1 - sin^2(x) to get rid of cosine. Simplify.

Answer 7

Oh you're right! Now I have to find the POI but how does sin(x)=-8/3 ?

Answer 8

I mean how do you find x with that?

Answer 9

|sin(x)| <= 1. So it cannot be -8/3

Answer 10

Hmm well I'm confused...wait it says I have to determine concavity at x=pi so does that mean when f"(x)=-3sinx-8 I plug in pi?

Answer 11

Why didn't you include that you have to determine concavity at ** x = pi ** in the first place when you originally posted the question?

Anyway, find f''(pi).

Answer 12

Yeah well it was on the next page of the packet so...yeah sorry about that. Anyways thanks for your help I got the answer and understand how I got it.

Answer 13

Alright.