Answer
$(1,-2)$ and $(-1,2)$
Work Step by Step
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 $(-1,2)$
