Question

Find the value of $x$ such that $f\left(x\right)=2x^3-15x^2+36x+1$ is an increasing function.

Answer 1

$f\left(x\right)=2x^3-15x^2+36x+1$
$\therefore f^{\prime }\left(x\right)=6x^2-30x+36$
$\therefore f^{\prime }\left(x\right)=6(x^2-5+6)$
i,e. $f^{\prime }\left(x\right)=6(x-3)(x-2)$
$\because f\left(x\right)$ is increasing function.
$\therefore f^{\prime }\left(x\right)\ge 0$
$6(x-3)(x-2)\ge 0$
$\therefore x>3$ or $x<2$
$\therefore f\left(x\right)$ is increasing $x>3$ or $x<2$.