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)

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,

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

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