Rolle’s theorem is a special case of

The correct option is Lagrange’s mean value theorem.

To understand why Rolle's theorem is a special case of Lagrange's Mean Value Theorem (LMVT), let us analyze the statements of both theorems.

Lagrange’s Mean Value Theorem:
If a function $f\left(x\right)$ is:
1. Continuous on the closed interval $[a,b]$, and
2. Differentiable on the open interval $(a,b)$,
then there exists at least one point $c$ in the open interval $(a,b)$ such that:

$$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$$

This formula represents the slope of the tangent line at $x=c$ being equal to the slope of the secant line passing through the endpoints $(a,f(a\left)\right)$ and $(b,f(b\left)\right)$.

Rolle’s Theorem:
Rolle's theorem adds a third specific condition to the first two:
3. The function values at the endpoints are equal, i.e., $f\left(a\right)=f\left(b\right)$.

If we apply this special condition $f\left(a\right)=f\left(b\right)$ to the equation in Lagrange's Mean Value Theorem, we get:

$$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(b\right)}{b-a}=0$$

Thus, $f^{\prime }\left(c\right)=0$, which is the exact conclusion of Rolle’s theorem.

Therefore, Rolle's theorem is a specific case of Lagrange's Mean Value Theorem where the secant line is horizontal (slope is zero).