Question

so, i'm finding the critical numbers of functions. The function is this: h(x)=sin^2 x+cos x and 0 12 years ago

Answer 1

Sorry, I can't help as I forgot most of my trignometry from 20 years ago.

Answer 2

yeahh. stupid Calculus.

Answer 3

I got all excited when it said my question was answered. dang.

Answer 4

a critical number is the value of x when derivative of the function is zero.
So, the derivative of h(x) = d/dx (sin^2x+cosx) = 0.
I think d/dx (sin^2x) = 2sinxcosx
I think d/dx cosx = -sinx

=> h'(x) = 2 sinx cosx - sinx = 0.
=> sinx(2cosx-1) = 0
=> critical points are value of x when sinx = 0
and 2 cosx - 1 = 0 or cos x = 1/2

Hope this helps or gets you on the right track.

Answer 5

Thank you! it did. the square trig function always throws me off. That's right! thanks!

Answer 6

x = 0 and x = 60?

Answer 7

i think it's x=0 degrees and x=90 degrees.

Answer 8

cos x = 1/2 when x = 60. So, I am not sure about 90.

Answer 9

yeah that's true..

Answer 10

sorry. i was looking at the wrong thing. you are right. so 0 to 2 pii is the whole circle right? so both answers will work as critical points?

Answer 11

Yes.

Answer 12

genius.