## Calculus (3rd Edition)

$x=2$, $x=-2$
The tangent line is horizontal when the slope is 0. This implies that $f'(x)=0$. $f'(x)=\frac{d}{dx}(12x-x^{3})=12\times\frac{d}{dx}(x)-\frac{d}{dx}(x^{3})$ $=(12\times1)-3x^{3-1}=12-3x^{2}$ $12-3x^{2}=0$ when $x=2$ or $x=-2$