Prove that the following function is increasing on $r$ ?
i. $f(x)=3 x^{5}+40 x^{3}+240 x$
ii. $f(x)=4 x^{3}-18 x^{2}+27 x-27$
(i) we have
$f(x)=3 x^{5}+40 x^{3}+240 x$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=15 \mathrm{x}^{4}+120 \mathrm{x}^{2}+240$
$=15\left(x^{4}+8 x^{2}+16\right)$
$=15\left(x^{2}+4\right)^{2}$
Now,
$x \in R$
$\Rightarrow\left(\mathrm{x}^{2}+4\right)^{2}>0$
$\Rightarrow 15\left(\mathrm{x}^{2}+4\right)^{2}>0$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0$
Hence, $f(x)$ is an increasing function for all $x$
(ii) we have
$f(x)=4 x^{3}-18 x^{2}+27 x-27$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=12 \mathrm{x}^{2}-36 \mathrm{x}+27$
$=12 \mathrm{x}^{2}-18 \mathrm{x}-18 \mathrm{x}+27$
$=3(2 \mathrm{x}-3)^{2}$
Now,
$x \in R$
$\Rightarrow(2 x-3)^{2}>0$
$\Rightarrow 3(2 x-3)^{2}>0$
$\Rightarrow f^{\prime}(x)>0$
Hence, $f(x)$ is an increasing fuction for all $x$