1. The problem asks for the number of critical points of the function $f(x) = x^3 - 3x$ defined on the interval $]-1,4[$. 2. Critical points occur where the derivative $f'(x)$ is zero or undefined. First, find the derivative: $$f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3$$ 3. Set the derivative equal to zero to find critical points: $$3x^2 - 3 = 0$$ $$x^2 = 1$$ $$x = \pm 1$$ 4. Since the domain is $]-1,4[$, the point $x = -1$ is not included (open interval), but $x = 1$ is included. 5. Check if $x = -1$ is a critical point within the domain: it is excluded, so not counted. 6. However, the function is differentiable everywhere, so no other critical points exist. 7. But note that the problem's graph description mentions three critical points: a local maximum near $x = -1$, a local minimum near $x = 1$, and an inflection point at $x=0$. 8. Inflection points are where the second derivative changes sign, not critical points. So $x=0$ is not a critical point. 9. Since $x=-1$ is not in the domain, only $x=1$ is a critical point inside the domain. 10. Therefore, the number of critical points of $f$ on $]-1,4[$ is 1. **Final answer:** (b) 1