Show that the function f(x) = sin(2x + π/4) is decreasing on (3π/8, 5π/8)

Question

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

Answer 1

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,

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

hence, Condition for f(x) to be decreasing

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