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

Given $y=6x^2-x^3$ The critical point occurs where the first derivative is zero: so $\frac{dy}{dx}=\frac{d{(6x^2-x^3)}}{dx}$ ${\frac{dy}{dx}}=12x-3x^2$ so the critical point is ${\frac{dy}{dx}}=0$ $f'(x)=12x-3x^2=0$ $x=0, x=4$ the critical value of y can be obtained by putting in the value of x: $y=6\times0^2-0=0$ $y=6\times4^2-4^3=32$ thus critical points are (0,0) and (4,32)