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# Prove that the following function is increasing on r ?

Question:

Prove that the following function is increasing on $r$ ?

i. $f(x)=3 x^{5}+40 x^{3}+240 x$

ii. $f(x)=4 x^{3}-18 x^{2}+27 x-27$

Solution:

(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 $x$