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$f(x)= 9\cos^2(x) - 18\sin(x),\quad 0 \le x \le 2\pi$

(a) Find the interval on which $f$ is increasing.

I answered: $(0, \frac {3\pi}{2})$

(b) Find the interval on which $f$ is decreasing.

I answered: $(\frac {3\pi}{2}, 2\pi)$

(c) Find the local minimum and maximum of $f$.

First off, am I correct with a and b? Second, I am getting really weird values when I test values on my derivate. For example: $\frac {\pi}{2}$ entered into the derivative returns $0$. Whereas if I was to enter $\pi$ I get a positive value($18$). How do I find the minimum and maximum of this function?

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    So I solved for the derivative $f'(x) = -18cos(x)(sin(x) + 1)$ and found what values made it equal to $0$, ($\frac {3\pi}{2}$), and then I tested for values less than $\frac {3\pi}{2}$. Then for values greater than $\frac {3\pi}{2}$. I found that some of the values less than $\frac {3\pi}{2}$ were increasing and that some of the values greater than $\frac {3\pi}{2}$ were decreasing.2012-11-11

2 Answers 2

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Apparently you're given the definition domain of the function to be $\,[0,2\pi]\,$, so:

$f'(x)=-18\cos x\sin x-18\cos x=-18\cos x(\sin x+1)\geq 0\Longleftrightarrow \cos x\leq 0\Longleftrightarrow$

$\Longleftrightarrow x\in \left[\frac{\pi}{2}\,,\,\frac{3\pi}{2}\right]$

You could see you're wrong since

$f'\left(\frac{\pi}{4}\right)=-18\frac{1}{\sqrt 2}\left(\frac{1}{\sqrt 2}+1\right)<0$

which is impossible if the function's ascending.

Added on request of the OP: You have right the second derivative, but then:

$f''(x)=18(\sin^2x+\sin x+\sin^2x -1)=0\Longleftrightarrow 2\sin^2x+\sin x -1=0\Longleftrightarrow$

$\Longleftrightarrow 2(\sin x+1)\left(\sin x-\frac{1}{2}\right)=0$

Can you take it from here?

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    Yes...and what about $\,\frac{3\pi}{2}\,$? It is not an inf. point, but did you check this?2012-11-11
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Solution not using calculus, so might not be acceptable for homework.

$ f(x) = 9\cos^2x - 18\sin x $ $ = 9(1-\sin^2 x) - 18 \sin x$ $ = 18 -9(\sin^2 x + 2 \sin x + 1) $ $ = 18 - 9(\sin x + 1)^2$

Now $\sin x$ increases in $[0, \pi/2]$ and $[3\pi/2, 2 \pi]$ so $(1 + \sin x )^2$ will increase and $f(x)$ will decrease. This leave us with $[\pi/2, 3\pi/2]$ where $f(x)$ increases.

Local maximum and minimum will occur when $f(x)$ will go from increasing to decreasing or vice versa or at the end points of the domain.

This gives $x = 0, \pi/2, 3\pi/2$ and $2 \pi$ as possible points to check for.

It is easy to check $f(x) = 9$ at end points, -18 at $\pi/2$ and 18 at $3\pi/2$.