Consider the equation below. f(x) = 3 cos^2x − 6 sin x, 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) b)Find the interval on which f is decreasing. (Enter your answer in interval notation.) c)(b) Find the local minimum and maximum values of f. d)(c) Find the inflection points. e)Find the interval on which f is concave up. (Enter your answer in interval notation.) f)Find the interval on which f is concave down. (Enter your answer in interval notation.) Please help me work this out!

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Consider the equation below. f(x) = 3 cos^2x − 6 sin x, 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) b)Find the interval on which f is decreasing. (Enter your answer in interval notation.) c)(b) Find the local minimum and maximum values of f. d)(c) Find the inflection points. e)Find the interval on which f is concave up. (Enter your answer in interval notation.) f)Find the interval on which f is concave down. (Enter your answer in interval notation.) Please help me work this out!

Mathematics
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\[f(x)=3\cos^2(x)-6\sin(x)\] right?
yes
did you take the derivative as a first step?

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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?
once you factor out the \(-6\cos(x)\) it is going to be real easy to see where the derivative is positive and negative
-6cos(x)(sin(x)+1)?
yeah and forturnately for you, since \(\sin(x)\geq -1\) always, you know \(\sin(x)+1\geq 0\) so you can ignore that part when you check tor the sign of the derivative
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
it is clear that in order to find the interval over which the function is increasing (decreasing) your only job is to find the interval over which the derivative is positive (negative) yes?
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
that means when you are trying to find if the derivative is positive or negative, you can ignore that factor entirely (because it is never negative) and concentrate only on \(-6\cos(x)\)
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 \((\frac{3\pi}{2},2\pi)\)
since your cosine has a \(-6\) in front of it, it will be negative where cosine is positive, and vice versa
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

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