The value of c in Rolle’s theorem for the function f(x) = x³ – 3x in the interval [o, √3] is

The correct option is 1.

Rolle's theorem states that if a real-valued function $f\left(x\right)$ is:
1. Continuous on a closed interval $[a,b]$,
2. Differentiable on the open interval $(a,b)$, and
3. $f\left(a\right)=f\left(b\right)$,
then there exists at least one value $c$ in the open interval $(a,b)$ such that the derivative at that point is zero, i.e., $f\text{'}\left(c\right)=0$.

Let us verify the conditions of Rolle's theorem for the function $f\left(x\right)=x^3-3x$ on the interval $[0,\sqrt{3}]$:
- Since $f\left(x\right)$ is a polynomial function, it is continuous on $[0,\sqrt{3}]$ and differentiable on $(0,\sqrt{3})$.
- Check the values of the function at the boundaries of the interval:

$$f\left(0\right)=0^3-3\left(0\right)=0$$

$$f\left(\sqrt{3}\right)=(\sqrt{3}{)}^3-3(\sqrt{3})=3\sqrt{3}-3\sqrt{3}=0$$

Since $f\left(0\right)=f\left(\sqrt{3}\right)=0$, all three conditions of Rolle's theorem are satisfied.

Now, we find the derivative $f\text{'}\left(x\right)$:

$$f\text{'}\left(x\right)=\frac{d}{dx}(x^3-3x)=3x^2-3$$

To find the value of $c$, we set $f\text{'}\left(c\right)=0$:

$$3c^2-3=0$$

$$3c^2=3$$

$$c^2=1$$

$$c=\pm 1$$

According to Rolle's theorem, the value $c$ must lie within the open interval $(0,\sqrt{3})$.
Comparing our two potential values:
- $c=-1$ does not belong to the interval $(0,\sqrt{3})$.
- $c=1$ lies within the interval $(0,\sqrt{3})$ since $0<1<\sqrt{3}\approx 1.732$.

Therefore, the value of $c$ is 1.