The number of points on the curve $y=x^3-4x$ at which tangent is parallel to x-axis

Given curve is $y=x^3-4x$.

Finding the slope of tangent by differentiating the curve with respect to $x$ we get,

$\frac{dy}{dx}=3x^2-4$

For the tangent to parallel to $x$- axis, the slope should be zero that is,

$$\begin{gathered} \frac{dy}{dx}=0 \\ 3x^2-4=0 \\ x=\pm \frac{2}{\sqrt{3}} \end{gathered}$$

For $x=\frac{2}{\sqrt{3}}$ we have $y=0$.

For $x=-\frac{2}{\sqrt{3}}$ we have $y=0$.

Therefore, there are $2$ points at which the tangent is parallel to $x$-axis which are $\left(\frac{2}{\sqrt{3}},0\right),\left(-\frac{2}{\sqrt{3}},0\right)$.