Answer :

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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