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Q.
For all real values of x, increasing function f(x) is:
Application of Derivatives
Solution:
Note: A function f (x) is s.t.b increasing function if f '(x) > 0
(i) Now, if f (x) = $\frac{1}{x} \, \Rightarrow \, f'(x) = \frac{-1}{x^2}$ the function becomes decreasing for x > 0
(ii) For, $f(x) = x^3 \, \Rightarrow \, f'(x) = 3x^2$
$3x^2$ is always > 0 for any real value of x
(iii) For $f (x) = x^2 \, \Rightarrow f '(x) = 2x $
which clearly shows that f (x) is a decreasing function for x < 0