Question

Find the points of maximum and minimum and the intervals of monotonicity of the following functions $f\left(x\right)=(x-1)e^{3x}$

Answer 1

$f\left(x\right)=(x-1)e^{3x}$

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

and decreasing $f^{\prime }\left(x\right)<0$

$\Rightarrow f^{\prime }\left(x\right)=(x-1)\frac{d}{dx}\left(e^{3x}\right)+e^{3x}\frac{d}{dx}(x-1)$

$=(x-1)\frac{d}{dx}\cdot 3e^{3x}+e^{3x}$

$=e^{3x}(3x-3+1)$

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

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

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

$\Rightarrow (3x-2)>0$

$x>\frac{2}{3}xϵ(\frac{2}{3},\infty )$

and for decreasing $f^{\prime }\left(x\right)<0$

$xϵ(-\infty ,\frac{2}{3})$

for maxima and minima $f^{\prime }\left(x\right)=0$

$e^{3x}(3x-2)=0$

$x=\frac{2}{3}$

$f^{\prime \prime }\left(x\right)=e^{3x}\frac{d}{dx}(3x-2)+(3x-2)\frac{d}{dx}\left(e^{3x}\right)$

$=e^{3x}\left(3\right)+(3x-2)3\cdot e^{3x}$

$=3e^{3x}[1+3x-2]$

$=3e^{3x}[3x-1]$

at $x=\frac{2}{3}$

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

So $x=\frac{2}{3}$ is point of local maxima.