Show that f(x) = cos x is a decreasing function on (0, π), increasing in (–π, 0) and neither increasing nor decreasing in (–π, π).

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 , 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) = cos x

⇒

⇒ f’(x) = –sinx

Taking different region from 0 to 2π

a) let

⇒ sin(x) > 0

⇒ –sinx < 0

⇒ f’(x) < 0

Thus f(x) is decreasing in

b) let

⇒ sin(x) < 0

⇒ –sinx > 0

⇒ f’(x) > 0

Thus f(x) is increasing in

Therefore, from above condition we find that

⇒ f(x) is decreasing in and increasing in

Hence, condition for f(x) neither increasing nor decreasing in (–π,π)

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