the length of a longest interval in which the function :
3sinx-4(sinx)^3 is increasing
a) pi/3
b)pi/2
c)3pi/2
d) pi
Your function is clearly $\displaystyle f(x)=sin3x$
$\displaystyle f'(x)=3cos3x>0$ for $\displaystyle 3x\in(-\frac{\pi}{2},\frac{\pi}{2})$ i.e for $\displaystyle x\in(-\frac{\pi}{6},\frac{\pi}{6}) $
Therefore the length of the largest interval in which $\displaystyle f(x)$ is increasing is $\displaystyle =\frac{\pi}{6}-(-\frac{\pi}{6})$ = $\displaystyle \frac{\pi}{3}$
A "function" should have a name attached. (What has been posted is just an "expression".) So let's insert the name they omitted when they composed this exercise:
. . . . .$\displaystyle f(x)\, =\, 3\sin{(x)}\, -\, 4\sin^3{(x)}$
As you saw in the previous reply, the customary way to answer questions of this sort is to differentiate:
. . . . .$\displaystyle f'(x)\, =\, 3\cos{(x)}\, -\, 12\sin^2{(x)}\cos{(x)}$
. . . . . . . . .$\displaystyle =\, 3\cos{(x)}\left(1\, -\, 4\sin^2{(x)}\right)$
Then one would set the derivative equal to zero, and solve for the critical points:
. . . . .$\displaystyle 3\cos{(x)}\, =\, 0\, \mbox{ or }\, 1\, -\, 4\sin^2{(x)}\, =\, 0$
. . . . .$\displaystyle \cos{(x)}\, =\, 0\, \mbox{ or }\, \sin^2{(x)}\, =\, \frac{1}{4}$
...and so forth.
However, you posted this exercise to the pre-calculus category, so you probably haven't learned anything about derivatives yet. Please reply with an explanation of whatever method you are supposed to be using to find the answer.
When you reply, please include a clear listing of your steps and reasoning so far. Thank you!