Given:- Function \(\text{f(x)}=\text{sin} \bigg(2x + \frac{\pi}{4} \bigg)\)
Theorem:- Let f be a differentiable real function defined on an open interval (a, b).
(i) If f’(x) > 0 for all x ∈ (a, b), then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all x ∈ (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’(x)
(iii) Put f’(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,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=16330816835852401611)
as here \(2x + \frac{\pi}{4}\) lies in 3rd quadrant
![](https://www.sarthaks.com/?qa=blob&qa_blobid=2472618304798602057)
hence, Condition for f(x) to be decreasing
Thus f(x) is decreasing on interval \(\big(\frac{3\pi}{8},\frac{5\pi}{8}\big)\)