Finding Critical Numbers Part IV

[Hckyplayer8]

Find the critical numbers for f(x)=sin²x + cos x in the interval [0,pi] and classify each one as a local max, min or neither.

First we must find the derivative, thus it looks like we must use the chain rule which I admit I am still terrible at. Providing this is a chain rule question, well...at least I can identify when I need to use it.

f ' (x) = (cos x²) (2x) - (sin x) ?

 

[Hckyplayer8]

Then providing that is correct, we set that to 0 to find out the critical numbers. But I'm not too sure how.

 

[Harry_the_cat]

Your derivative of the first term is incorrect.
Think of sin²x as (sin x)² and use the chain rule. Try again.

 

[HallsofIvy]

In the statement of the problem you have \(\displaystyle sin^2(x)\) but in your derivative you treat it as \(\displaystyle sin(x^2)\). Which is it?

 

[Hckyplayer8]

In the statement of the problem you have \(\displaystyle sin^2(x)\) but in your derivative you treat it as \(\displaystyle sin(x^2)\). Which is it?

Oops. It is sin²(x)

 

[Hckyplayer8]

After going back an redoing the first half of the derivative, it should be f ' (x) = sin (2x) - sin x

 

[Harry_the_cat]

After going back an redoing the first half of the derivative, it should be f ' (x) = sin (2x) - sin x

Yes it is but probably best to leave it as 2 sin x cos x - sin x. Then it can be factorised.

 

[Hckyplayer8]

Yes it is but probably best to leave it as 2 sin x cos x - sin x. Then it can be factorised.

Okay. Factored is (sin x)(2 cos x-1). Now I set that to zero?

 

[Harry_the_cat]

Okay. Factored is (sin x)(2 cos x-1). Now I set that to zero?

Yes to find the critical values.