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\[f(x)=3\cos^2(x)-6\sin(x)\] right?

yes

did you take the derivative as a first step?

and if so, what did you get?

-6cos(x)sin(x)-6cos(x)

that looks good
i guess the next step is to factor

you got that?

-6cos(x)(sin(x)+1)?

in other words, the sign of the derivative is completely dependent on the sign of \(-6\cos(x)\)

you got that?

no, im confused as to what my answer would be

ok lets go slow

yes

ok and your derivative isi
\[-6\cos(x)\left(\sin(x)+1\right)\]

so the question is, over what interval is the derivative positive

so i would have to graph it to see it clearly right?

you have two factors \[-6\cos(x)\] and \[\sin(x)+1\]
no you do not have to graph it

is it clear that
\[\sin(x)+1\geq 0\] for any value of \(x\)?

yes

do you know over what intervals between \(0\) and \(2\pi\) that cosine is positive?

i cant remember

then i guess i have to tell you

it is positive on \((0,\frac{\pi}{2})\) that is on the right sides of the unit circle

negative on \((\frac{\pi}{2},\frac{3\pi}{2})\) that is the left side

and positive again on \((\frac{3\pi}{2},2\pi)\)

okay so for a) it would be pi/2,3pi/2 ?

increasing on that interval, yes

and of course decreasing on the remaining intervals

to find the max and min i would plug in to the equation pi/2 and then 3pi/2?

yes

in to the original function of course, not the derivative

got 6 and -6

looks good

thank you so much!

yw