Show that the function

Solution:

Given:- Function $\mathrm{f}(\mathrm{x})=\sin \left(2 \mathrm{x}+\frac{\pi}{4}\right)$

Theorem:- Let $\mathrm{f}$ be a differentiable real function defined on an open interval $(\mathrm{a}, \mathrm{b})$.

(i) If $f^{\prime}(x)>0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.

Here we have,

$\mathrm{f}(\mathrm{x})=\sin \left(2 \mathrm{x}+\frac{\pi}{4}\right)$

$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin \left(2 \mathrm{x}+\frac{\pi}{4}\right)\right\}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\cos \left(2 \mathrm{x}+\frac{\pi}{4}\right) \times 2$

$\Rightarrow \mathrm{f}(\mathrm{x})=2 \cos \left(2 \mathrm{x}+\frac{\pi}{4}\right)$

Now, as given

$x \in\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$

$\Rightarrow \frac{3 \pi}{8}

$\Rightarrow \frac{3 \pi}{4}<2 x<\frac{5 \pi}{4}$

$\Rightarrow \pi<2 x+\frac{\pi}{4}<\frac{3 \pi}{2}$;

as here $2 x+\frac{\pi}{4}$ lies in $3^{\text {rd }}$ quadrant

$\Rightarrow \cos \left(2 x+\frac{\pi}{4}\right)<0$

$\Rightarrow 2 \cos \left(2 x+\frac{\pi}{4}\right)<0$

$\Rightarrow f^{\prime}(x)<0$

hence, Condition for $f(x)$ to be decreasing

Thus $f(x)$ is decreasing on interval $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$