1. **Problem statement:** Prove Lagrange's theorem (Mean Value Theorem), which states: If a function $f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists at least one point $c \in (a,b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$ 2. **Step 1: Construct an auxiliary function.** Define $$F(x) = f(x) - \lambda x$$ where $\lambda$ is a constant to be determined such that $F(a) = F(b)$. 3. **Step 2: Determine $\lambda$.** Using $F(a) = F(b)$, $$f(a) - \lambda a = f(b) - \lambda b$$ Rearranging, $$\lambda (b - a) = f(b) - f(a)$$ So, $$\lambda = \frac{f(b) - f(a)}{b - a}.$$ 4. **Step 3: Verify conditions for Rolle's theorem.** Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $F(x)$ differs only by a linear term, $F$ is also continuous on $[a,b]$ and differentiable on $(a,b)$. Also, by construction, $$F(a) = F(b).$$ So $F$ satisfies the hypotheses of Rolle’s theorem. 5. **Step 4: Apply Rolle's theorem.** By Rolle's theorem, there is $c \in (a,b)$ such that $$F'(c) = 0.$$ Since $$F'(x) = f'(x) - \lambda,$$ we have $$0 = F'(c) = f'(c) - \lambda \Rightarrow f'(c) = \lambda = \frac{f(b) - f(a)}{b - a}.$$ 6. **Step 5: Conclusion.** We have shown there exists $c \in (a,b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a},$$ which is Lagrange's mean value theorem. 7. **Geometrical meaning:** If $A=(a,f(a))$ and $B=(b,f(b))$ are points on the curve, the chord $AB$ has slope $\frac{f(b)-f(a)}{b-a}$. The theorem guarantees at least one point $C=(c,f(c))$ between $A$ and $B$ where the tangent line to the curve is parallel to the chord $AB$ (meaning the derivative at $c$ equals the slope of $AB$).