Question

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

Answer 1

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

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

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

$=3x^2-6x+3+1$

$=3(x^2-x+1)+1$

$=3(x-1{)}^2+1$

As square is a positive number.

The value of $f^{\prime }\left(x\right)$ will always be positive for real number.