Show that f(x) = x – sin x is increasing for all x ϵ R.

Given:- Function f(x) = x – sin 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) = x – sin x

⇒

⇒ f’(x) = 1 – cos x

Now, as given

x ϵ R

⇒ –1 < cosx < 1

⇒ –1 > cos x > 0

⇒ f’(x) > 0

hence, Condition for f(x) to be increasing

Thus f(x) is increasing on interval x ∈ R

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