## University Calculus: Early Transcendentals (3rd Edition)

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)