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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ How have you arrived at that answer? $\endgroup$ – mfl Mar 29 '15 at 21:42
  • 1
    $\begingroup$ But what did you do for the first part? $\endgroup$ – danimal Mar 29 '15 at 21:45
  • 1
    $\begingroup$ 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


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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.