## Thomas' Calculus 13th Edition

$(1,-2)$ and $(-1,2)$
Given that $f(x)=x^3-3x$, the function will have a horizontal tangent when the derivative is equal to 0, since the slope will indicate that there is no change in the y value of the tangent when x changes. $f'(x)$ is found using the power rules: $f'(x)=3x^2-3$ We let $f'(x)$ be equal to zero. $3x^2-3=0$ $x^2=1$ $x=+1$ or $x=-1$ $f(1)=-2$ and $f(-1)=2$ Thus, there exist two different maxima and/or minima on the function where both points have horizontal tangents. $(1,-2)$ and $(-1,2)$