Given:- Function f(x) = cos x
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,
Taking different region from 0 to 2π
Thus f(x) is decreasing in (0, π)
Thus f(x) is increasing in (-π, 0)
Therefore, from above condition we find that
⇒ f(x) is decreasing in (0, π) and increasing in (-π, 0)
Hence, condition for f(x) neither increasing nor decreasing in (–π, π)