Question

Let a function $f\left(x\right)=\frac{x^3}{3}-2x^2+3x$ on the set $A=\{x|x^2+8\le 6x\},$ then the minimum value of $f$ is

Answer 1

Given : $f\left(x\right)=\frac{x^3}{3}-2x^2+3x$
$\Rightarrow f^{\prime }\left(x\right)=x^2-4x+3=(x-1)(x-3)$
$\Rightarrow f^{\prime }\left(x\right)=0,$ gives $x=1,3$
now, set $A=\{x|x^2-6x+8\le 0\}$
$\Rightarrow \text{set}A=[2,4]$
So, critical points for $f\left(x\right)$ is $x=2,3,4$ and
$f\left(2\right)=\frac{2}{3}f\left(3\right)=0f\left(4\right)=\frac{4}{3}$
So, $f_{min}=0$