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Find the intervals in which f(x) = sin x – cos x, where 0 < x < 2π is increasing or decreasing.
Given:- Function f(x) = sin x – cos x, 0 < x < 2π
Theorem:- Let f be a differentiable real function defined on an open interval (a,b).
(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)
(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,
f(x) = sin x – cos x
⇒ f’(x) = cos x + sin x
For f(x) lets find critical point, we must have
⇒ f’(x) = 0
⇒ cos x + sin x = 0
⇒ tan(x) = –1
Here these points divide the angle range from 0 to 2∏ since we have x as angle
clearly, f’(x) > 0 if
and f’(x) < 0 if
Thus, f(x) increases on
and f(x) is decreasing on interval