Question

Show that the function $f\left(x\right)=\sin (2x+\pi /4)$ is decreasing on $(3\pi /8,5\pi /8)$

Answer 1

We have

$f\left(x\right)=\sin (2x+\pi /4)\Rightarrow f^{\prime }\left(x\right)=2\cos (2x+\pi /4)$

Now, $x\in (3\pi /8,7\pi /8)\Rightarrow 3\pi /8<x<5\pi /8\Rightarrow 3\pi /4<2x<5\pi /4\Rightarrow \pi <2x+\pi /4<\frac{3\pi }{2}$

$\therefore$ $2\cos (2x+\pi /4)<0\Rightarrow f^{\prime }\left(x\right)<0$

So, $f\left(x\right)$ is decreasing on $(3\pi /8,7\pi /8)$