# Verify Lagrange's mean value theorem for the following function: $f(x)=x^2-2x+3,\: for \: [4,6]$

Toolbox:
• Lagrange's Mean Value Theorem :
• Let $f(x)$ be a real valued function that satisfies the following conditions.
• (i) $f(x)$ is continuous on the closed interval $[a,b]$
• (ii) $f(x)$ is differentiable in the open interval $(a,b)$
• (iii) $f(a)=f(b)$
• Then there exists atleast one value $c \in (a,b)$ such that $f'(c)=0$
• $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
Given :$f(x)=x^2-2x+3$ in the interval [4,6]
We know that a polynomial function is continuous everywhere and also differentiable.
So $f(x)$ being a polynomial is continuous and differentiable on (4,6)
So there must exist at least one real number $c\in (4,6)$ such that
$f'(c)=\large\frac{f(6)-f(4)}{6-4}$
Step 2:
$f(x)=x^2-2x+3$
$f(6)=6^2-2(6)+3=27$
$f(4)=4^2-2(4)+3=11$
$f'(x)=2x-2$
$f'(c)=2c-2$
$\therefore 2c-2=\large\frac{27-11}{2}$
$2c-2=\large\frac{16}{2}$
$2c-2=8$
$2c=10$
$c=5$
$c\in (4,6)$
Hence Lagrange's Mean Value theorem is verified.