The Full Form of LMVT is Lagrange’s mean value theorem.

Lagrange’s mean value theorem (MVT) states that if a function $f\left(x\right)$ is continuous on a closed interval $\left[a,b\right]$ and differentiable on the open interval $\left(a,b\right),$ then there is at least one point $x=c$ on this interval, such that

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

This theorem (also known as First Mean Value Theorem) allows to express the increment of a function on an interval through the value of the derivative at an intermediate point of the segment.

Proof.

Consider the auxiliary function

$$F\left(x\right)=f\left(x\right)+\lambda x.$$

We choose a number $\lambda$ such that the condition $F\left(a\right)=F\left(b\right)$ is satisfied. Then

$$f\left(a\right)+\lambda a=f\left(b\right)+\lambda b,\Rightarrow f\left(b\right)-f\left(a\right)=\lambda \left(a-b\right),\Rightarrow \lambda =-\frac{f\left(b\right)-f\left(a\right)}{b-a}.$$

As a result, we have

$$F\left(x\right)=f\left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}x.$$

The function $F\left(x\right)$ is continuous on the closed interval $\left[a,b\right],$ differentiable on the open interval $\left(a,b\right)$ and takes equal values at the endpoints of the interval. Therefore, it satisfies all the conditions of Rolle’s theorem. Then there is a point $c$ in the interval $\left(a,b\right)$ such that

$$F^{\prime }\left(c\right)=0.$$

It follows that

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

or

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