Question

Show that $f\left(x\right)=x^3-6x^2+18x+5$ is an increasing function for all $xϵR$.

Answer 1

We have,

$f\left(x\right)=x^3-6x^2+18x+5$

We need to show $f\left(x\right)$ is strictly increasing on $R$.

i.e. we show that $f^{\prime }\left(x\right)>0$

Then,

$f^{\prime }\left(x\right)=3x^2-12x+18+0$

$=3(x^2-4x+6)$

$=3(x^2-4x+4+2)\text{on}\text{completing}\text{square}$

$\text{=3}(x^2-4x+4)+6$

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

We know that,

Square of any number is always $\left(\text{+ive}\right)$

$\Rightarrow 3{(x-2)}^2+6>0$

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

Here $f^{\prime }\left(x\right)>0$for any value of $$x$$

$\therefore f\left(x\right)$ is strictly increasing on $R.$