Note: A function f (x) is s.t.b increasing function if f '(x) > 0
(i) Now, if f (x) = x1⇒f′(x)=x2−1 the function becomes decreasing for x > 0
(ii) For, f(x)=x3⇒f′(x)=3x2 3x2 is always > 0 for any real value of x
(iii) For f(x)=x2⇒f′(x)=2x
which clearly shows that f (x) is a decreasing function for x < 0