Answer
critical points are (1,0), (2,1) and (3,0)
Work Step by Step
Given $y=(x-3)^2(x-1)^2$
The critical points occur where the first derivative is zero:
so $\frac{dy}{dx}=\frac{d{(x-3)^2(x-1)^2}}{dx}$
${\frac{dy}{dx}}=2(x-3)(x-1)^2+2(x-3)^2(x-1)$
${\frac{dy}{dx}}=4(x-3)(x-2)(x-1)$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=4(x-3)(x-2)(x-1)=0$
$x=1,x=2,x=3$
the critical value of y can be obtained by putting in the value of x:
$y=(x-3)^2(x-1)^2=0$
$y=(x-3)^2(x-1)^2=1$
$y=(x-3)^2(x-1)^2=0$
thus the final critical points are (1,0), (2,1) and (3,0)
