Question

Show that the function f given by
$f\left(x\right)=x^3-3x^2+4x,xϵR$ is increasing on $R$

Answer 1

$f\left(x\right)=x^3-3x^2+4x$

Differentiating w.r.t $x$

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

$\Rightarrow f^{\prime }\left(x\right)=3x^2-6x+4$

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

$\Rightarrow f^{\prime }\left(x\right)=3(x^2-2x+1-1)+4$

$\Rightarrow f^{\prime }\left(x\right)=3\left[\right(x-1{)}^2-1]+4$

$\Rightarrow f^{\prime }\left(x\right)=(x-1{)}^2+1$

As $(x-1{)}^2\ge 0,\forall xϵR$

$\Rightarrow f^{\prime }\left(x\right)\ge 1>0$

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