# Verify Lagrange's theorem for the following function; $f(x)=2x^{3}+x^{2}-x-1\quad [0,2]$

Note: This is part 3rd of a 5 part question, split as 5 separate questions here.

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 $f'(c)=0$
$f(x)=2x^3+x^2-x-1 \quad [0,2]$
Step 1:
This is a poly nonomial function. It is continuous in $[0,2]$ and differentiable in $(0,2)$
Step 2:
To find $c \in (0,2)$ where $f'(c)=\large\frac{f(2)-f(0)}{2-0}$
$f'(x)=6x^2+2x-1 \qquad f'(c)=6c^2+2c-1$
$\large\frac{f(2)-f(0)}{2-0}=\large\frac{2(8)+4-2-1(-1)}{2}$$=9 Therefore 6c^2+2c-1=9 6c^2+2c-10=0 c=\large\frac{-2 \pm \sqrt {4 +240}}{12} \quad=\large\frac{-2 \pm 2\sqrt {61}}{12} \quad=\large\frac{-1 \pm \sqrt {61}}{6} The value \large\frac{-1 \pm \sqrt {61}}{6}$$ \in (0,2)$
The theorem is verified