anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[f(x)=3\cos^2(x)-6\sin(x)\] right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
did you take the derivative as a first step?

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anonymous
  • anonymous
and if so, what did you get?
anonymous
  • anonymous
-6cos(x)sin(x)-6cos(x)
anonymous
  • anonymous
that looks good i guess the next step is to factor
anonymous
  • anonymous
you got that?
anonymous
  • 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
  • anonymous
-6cos(x)(sin(x)+1)?
anonymous
  • 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
  • anonymous
in other words, the sign of the derivative is completely dependent on the sign of \(-6\cos(x)\)
anonymous
  • anonymous
you got that?
anonymous
  • anonymous
no, im confused as to what my answer would be
anonymous
  • anonymous
ok lets go slow
anonymous
  • 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
  • anonymous
yes
anonymous
  • anonymous
ok and your derivative isi \[-6\cos(x)\left(\sin(x)+1\right)\]
anonymous
  • anonymous
so the question is, over what interval is the derivative positive
anonymous
  • anonymous
so i would have to graph it to see it clearly right?
anonymous
  • anonymous
you have two factors \[-6\cos(x)\] and \[\sin(x)+1\] no you do not have to graph it
anonymous
  • anonymous
is it clear that \[\sin(x)+1\geq 0\] for any value of \(x\)?
anonymous
  • anonymous
yes
anonymous
  • 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
  • anonymous
do you know over what intervals between \(0\) and \(2\pi\) that cosine is positive?
anonymous
  • anonymous
i cant remember
anonymous
  • anonymous
then i guess i have to tell you
anonymous
  • anonymous
it is positive on \((0,\frac{\pi}{2})\) that is on the right sides of the unit circle
anonymous
  • anonymous
negative on \((\frac{\pi}{2},\frac{3\pi}{2})\) that is the left side
anonymous
  • anonymous
and positive again on \((\frac{3\pi}{2},2\pi)\)
anonymous
  • anonymous
since your cosine has a \(-6\) in front of it, it will be negative where cosine is positive, and vice versa
anonymous
  • anonymous
okay so for a) it would be pi/2,3pi/2 ?
anonymous
  • anonymous
increasing on that interval, yes
anonymous
  • anonymous
and of course decreasing on the remaining intervals
anonymous
  • anonymous
to find the max and min i would plug in to the equation pi/2 and then 3pi/2?
anonymous
  • anonymous
yes
anonymous
  • anonymous
in to the original function of course, not the derivative
anonymous
  • anonymous
got 6 and -6
anonymous
  • anonymous
looks good
anonymous
  • anonymous
thank you so much!
anonymous
  • anonymous
yw

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