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!

- anonymous

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- anonymous

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

- anonymous

yes

- anonymous

did you take the derivative as a first step?

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## More answers

- anonymous

and if so, what did you get?

- anonymous

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

- anonymous

that looks good
i guess the next step is to factor

- anonymous

you got that?

- anonymous

once you factor out the \(-6\cos(x)\) it is going to be real easy to see where the derivative is positive and negative

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you got that?

- anonymous

no, im confused as to what my answer would be

- anonymous

ok lets go slow

- anonymous

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?

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

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 \(-6\cos(x)\)

- anonymous

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

- anonymous

i cant remember

- anonymous

then i guess i have to tell you

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

since your cosine has a \(-6\) in front of it, it will be negative where cosine is positive, and vice versa

- anonymous

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

- anonymous

increasing on that interval, yes

- anonymous

and of course decreasing on the remaining intervals

- anonymous

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

- anonymous

yes

- anonymous

in to the original function of course, not the derivative

- anonymous

got 6 and -6

- anonymous

looks good

- anonymous

thank you so much!

- anonymous

yw

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