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I am having some trouble with the following question:

Find the critical points of the function and use the First Derivative Test to determine whether the critical point is a local minimum or maximum (or neither). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

Function: f(x) = 4 sin x cos x, on (0, π)

I was successfully able to get the local minimum and local maximum for this function which are:

3π/4 (local min)

π/4 (local max)

However, I have no idea what to do for the following:

Determine the intervals on which the function is increasing or decreasing. (Enter your answers using interval notation. Enter EMPTY or ∅ for the empty set.)

Any help is greatly appreciated.

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  • $\begingroup$ How have you got the local maximum/minimum? $\endgroup$ – mfl Mar 29 '15 at 21:41
  • $\begingroup$ Yes the local min is 3π/4 and the local max is π/4 $\endgroup$ – camrymps Mar 29 '15 at 21:41
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    $\begingroup$ How have you arrived at that answer? $\endgroup$ – mfl Mar 29 '15 at 21:42
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    $\begingroup$ But what did you do for the first part? $\endgroup$ – danimal Mar 29 '15 at 21:45
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    $\begingroup$ your function is $f = 4\sin x \cos x = 2 \sin 2x, 0 < x < \pi. $ $f$ is increasing on $(0,\pi/4) \cup (3\pi/4)$ and decreasing on $\pi/4, 3\pi /4)$ $\endgroup$ – abel Mar 29 '15 at 21:50
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Hint

You have

$$f'(x)=4(\cos^2 x-\sin^2 x).$$ To get the critical points you have to solve $f'(x)=0.$ You have done it and you have obtained $x=\pi/4$ and $x=3\pi/4.$ Now, you have to study the sign of $f$ on the intervals $(0,\pi/4),$ $(\pi/4,3\pi/4)$ and $(3\pi/4,\pi).$ ($0$ and $\pi$ because they are the extremes of the interval where the function is defined and $x\pi/4$ and $3\pi/4$ because they are the critical points.)

Remember that if $f'(x)>0,\: x\in (a,b)$ then $f$ is strictly increasing in $(a,b)$ and if $f'(x)<0,\: x\in (a,b)$ then $f$ is strictly decreasing in $(a,b).$ A local minimum is obtained when you change from an interval where the function is decreasing to an interval where it is increasing. It is similar for a local maximum. (This is the first derivative test.)

Can you finish?

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  • $\begingroup$ Okay so I drew a number line out and wrote down 0, π/4 and 3π/4 on it. Then I put arbitrary values above and below each of those (-5π/4, 5π/4. π/2) and then I plugged all of them into the first derivative to see which ones were positive and negative and this is what I got: the has a local max from 0 to 5π/4 and a local min from 5π/4 to 3π/4, but the rest of the function has neither local mins or local maxes. I wish I could post a picture of the number line I drew out. $\endgroup$ – camrymps Mar 29 '15 at 22:14
  • $\begingroup$ Yes. On each interval with $f'$ positive the function increases and on each interval with $f'$ negative the function decreases. (Note that $-5\pi/4<0$ and so you must consider, say, $\pi/3.$ Also, $\pi/2<3\pi/4,$ so use $4\pi/5.$) When $f'$ changes from positive to negative you have a local maximum and similar to local minimum. $\endgroup$ – mfl Mar 29 '15 at 22:19

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