Lagrange;s Mean Value Theorem Engineering Maths, Btech first year

Lagrange's Mean Value Theorem

If $f(x)$ is
a) continuous in [a,b]
b) differentiable in (a,b)
then there exists at least ne value $c\in (a,b)$ such that $$\frac{f(b)-f(a)}{b-a}=f^{\prime }(c)$$

Sample Problem

Verify Lagrange's Mean Value Theorem for $$f(x)=x^2\text{ in }(1,5)$$

Solution

$f(x)$ is continuous and differentiable in (1,5)
Here $a=1$ and $b=5$
$$\text{Now }{f}^{\prime }(x)=2x$$ So by Lagrange's Mean Value Theorem, $$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$ $$2c=\frac{5^2-1^2}{5-1}=\frac{25-1}{4}$$ $$\Rightarrow \boxed{{\displaystyle c=3}}$$