Prove that the function


Prove that the function $f(x)=x^{3}-6 x^{2}+12 x-18$ is increasing on $R$.


Given:- Function $f(x)=x^{3}-6 x^{2}+12 x-18$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$


(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $\mathrm{f}^{\prime}(\mathrm{x})>0$ 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,

$f(x)=x^{3}-6 x^{2}+12 x-18$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}-6 \mathrm{x}^{2}+12 \mathrm{x}-18\right)$

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

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=3\left(\mathrm{x}^{2}-4 \mathrm{x}+4\right)$

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

as given

$X \in R$


$\Rightarrow 3(x-2)^{2}>0$

$\Rightarrow f^{\prime}(x)>0$

Hence, condition for $f(x)$ to be increasing

Thus $f(x)$ is increasing on interval $x \in R$

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