x – x – x + f ′ ( x ) = x – x – = ( x – x – ) = ( x – ) ( x + ) Therefore, f ′ ( x ) = gives x = – , . The points x = – and x = divides the real line into three disjoint intervals, namely, (– ∞ , – ), (– , ) and ( , ∞ ). Fig . In the intervals (– ∞ , – ) and ( , ∞ ), f ′ ( x ) is positive while in the interval (– , ), f ′ ( x ) is negative.
Consequently, the function f is increasing in the intervals (– ∞ , – ) and ( , ∞ ) while the function is decreasing in the interval (– , ). However, f is neither increasing nor decreasing in R . Interval Sign of f ′ ( x ) Nature of function f (– ∞ , – ) (–) (–) > f is increasing (– , ) (–) (+) < f is decreasing ( , ∞ ) (+) (+) > f is increasing Example Find intervals in which the function given by f ( x ) = sin x , x ∈ , π is (a) increasing (b) decreasing. Solution We have f ( x ) = sin x f ′ ( x ) = 3cos x Therefore, f ′ ( x ) = gives cos x = which in turn gives (as x ∈ , π implies , ∈ ).
So and π . The point divides the interval , π into two disjoint intervals , and π π . Now, x > ′ for all