Answer
local minimum: $f(x)=0$ and $x=5$ and at $(0,0)$
local maximum: $f(x)=108$ and $x={3}$
Work Step by Step
Given $y=x^3(x-5)^2$
the critical point is where the first derivative is zero:
so $\frac{dy}{dx}=\frac{d{(x^3(x-5)^2)}}{dx}$
$\frac{dy}{dx}=\frac{{(3x^2(x-5)^2)-2x^3(x-5)}}{(x-5)^2}$
${\frac{dy}{dx}}={{5x^2(x-3)}}{(x-5)}$
so the critical point is: ${\frac{dy}{dx}}=0$
$f'(x)={{5x^2(x-3)}}{(x-5)}=0$
$x=0, x=3, x=5$
the critical value of y can be obtained by putting in the value of x:
at $x=5$: $y=x^3(x-5)^2=0$
at $x=3$: $y=x^3(x-5)^2=3^3(-2)^2=108$
at $x=0$: $y=x^3(x-5)^2=0$
Thus we have:
local minimum: $f(x)=0$ and $x=5$ and at $(0,0)$
local maximum: $f(x)=108$ and $x={3}$
