Question
Show that the function $\text{f}(\text{x})=\sin\Big(2\text{x}+\frac{\pi}{4}\Big)$ is decreasing on $\Big(\frac{3\pi}{8},\frac{5\pi}{8}\Big).$

Answer

$\text{f}(\text{x})=\sin\Big(2\text{x}+\frac{\pi}{4}\Big)$
$\text{f}'(\text{x})=2\cos\Big(2\text{x}+\frac{\pi}{4}\Big)$
Here,
$\frac{3\pi}{8}<\text{x}<\frac{5\pi}{8}$
$\Rightarrow\frac{3\pi}{4}<2\text{x}<\frac{5\pi}{4}$
$\Rightarrow\pi<2\text{x}+\frac{\pi}{4}<\frac{3\pi}{2}$
$\Rightarrow\cos\Big(2\text{x}+\frac{\pi}{4}\Big)<0$ $[\because$ Cos function is negative in third quadrent$]$
$\Rightarrow2\cos\Big(2\text{x}+\frac{\pi}{4}\Big)<0$
$\Rightarrow\text{f}'(\text{x})<0,\ \forall\ \text{x}\in\Big(\frac{3\pi}{8},\frac{5\pi}{8}\Big)$
So, f(x) is decreasing on $\Big(\frac{3\pi}{8},\frac{5\pi}{8}\Big).$

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