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Calculus1 9 Online
OpenStudy (anonymous):

sin^2x - cosx on the interval (0,2pi), im asked to find the intervals of increase and decrease...

OpenStudy (anonymous):

does f'(x)= 2sinxcosx + sinx?

OpenStudy (p0sitr0n):

yes its the f'(x)

OpenStudy (p0sitr0n):

when f'(x) positive - increase f'(x) negative - decrease f'(x) 0 = stays same

OpenStudy (anonymous):

so then i would have 2sinxcosx= -sinx because i have to set it equal to zero, so can i say that 2sinx=-tanx?

OpenStudy (p0sitr0n):

solve = 0 and study the sign of the f'(x)

OpenStudy (p0sitr0n):

i would divide it by sinx to get 2cosx=-1

OpenStudy (anonymous):

right, good thinking, so then cos x= -1/2 and x=2pi/3 and 5pi/3

OpenStudy (p0sitr0n):

cosx=-1/2 remembering the trig circle, -1/2 is 2pi/3 and 4pi/3

OpenStudy (anonymous):

oops 4pi/3

OpenStudy (p0sitr0n):

ya u got it. after u study the sign

OpenStudy (p0sitr0n):

and thats it

OpenStudy (anonymous):

ok thank you so much

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