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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeahh. stupid Calculus.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

x = 0 and x = 60?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah that's true..

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

genius.

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