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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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You are wrong on both (a) and (b). Can you explain how you got your answers? Your derivatives that you mentioned appear to be correct. Find min and max by setting the derivative equal to zero, like always (you already found one of them...). – Jonathan Christensen Nov 11 '12 at 3:11
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. – dsta Nov 11 '12 at 3:14
up vote 1 down vote accepted

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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What would the inflection points be for this function? I have $f''(x) = 18(sin^2(x) + sin(x) - cos^2(x))$. Thanks again! – dsta Nov 11 '12 at 3:27
I've added some stuff to my answer. If you have more questions perhaps it'd be a good idea to open a new thread. – DonAntonio Nov 11 '12 at 3:33
Inflection points are at x=$\frac {\pi}{6}$ and $\frac {5\pi}{6}$? – dsta Nov 11 '12 at 3:35
Yes...and what about $\,\frac{3\pi}{2}\,$? It is not an inf. point, but did you check this? – DonAntonio Nov 11 '12 at 3:38
Thank you so much! – dsta Nov 11 '12 at 3:40

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 $3\pi/2$.

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