# Mark the correct alternative in the following:

Question:

Mark the correct alternative in the following:

Let $f(x)=x^{3}-6 x^{2}+15 x+3 .$ Then,

A. $f(x)>0$ for all $x \in R$

B. $f(x)>f(x+1)$ for all $x \in R$

C. $f(x)$ in invertible

D. $f(x)<0$ for all $x \in R$

Solution:

Formula:- (i) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)>0$ for all $x \in(a, b)$

(ii) If $f(x)$ is strictly increasing function on interval $[a, b]$, then $f^{-1}$ exist and it is also a strictly increasing function

Given:- $f(x)=x^{3}-6 x^{2}+15 x+3$

$\frac{\mathrm{d}(f(x))}{d x}=3 x^{2}-12 x+15=f^{\prime}(x)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3(\mathrm{x}-2)^{2}+\frac{1}{3}$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3(\mathrm{x}-2)^{2}+\frac{1}{3}$

Therefore $f^{\prime}(x)$ will increasing

Also $f^{-1}(x)$ is possible

Therefore $f(x)$ is invertible function.