determine-the-intervals-on-which-the-function-fxsin2x-fx-sin2x-fxsin2x-is-increa

Question

Determine the intervals on which the function ﻿f(x)=sin⁡2(x) f(x) = \$$sin^2(x) f(x)=sin2(x)﻿ is $$increasing or decreasing on the interval ﻿[−π2,π2] \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] [−2π​,2π​]﻿.

Accepted Answer

Decreasing on ﻿(−π2,0)\left(-\frac{\pi}{2},0\right)(−2π​,0)﻿ Increasing on ﻿(0,π2)\left(0,\frac{\pi}{2}\right)(0,2π​)﻿

Answer

Increasing on ﻿(−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)(−2π​,2π​)﻿

Answer

Decreasing on (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Answer

Decreasing on ﻿(0,π2)\left(0,\frac{\pi}{2}\right)(0,2π​)﻿ Increasing on ﻿(−π2,0)\left(-\frac{\pi}{2},0\right)(−2π​,0)﻿

Determine the intervals on which the function f(x)=sin2(x) f(x) = \(\sin\)^2(x) f(x)=sin2(x) is increasing or decreasing on the interval [−π2,π2] \(\left\)[-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\]\right\)] [−2π,2π].

Determine the intervals on which the function $f(x) = \(\sin\)^2(x)$ is increasing or decreasing on the interval $\(\left\)[-\(\frac{\pi}{2}\), \(\frac{\pi}{2}\]\right\)]$ .