Question

Verify Lagrange's mean value theorem for $f\left(x\right)=2x^2-7x-10$ over $[2,5]$ and find $c$.

Answer 1

$f\left(x\right)=2x2-7x-10$ over $[2,5]$

We know that a polynomial function is continuous everywhere and also differentiable.

So $f\left(x\right)$ being a polynomial is continuous and differentiable on $(2,5)$

So there must exist at least one real number $c\in (2,5)$ such that

$f^{\prime }\left(c\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}$

$f\left(x\right)=2x2-7x-10$

$f\left(5\right)=252-7\times 5-10=50-35-10=5$

$f\left(2\right)=222-7\times 2-10=8-14-10=-14$

$f^{\prime }\left(x\right)=4x-7$

$f^{\prime }\left(c\right)=4c-7$

$\therefore 4c-7=\frac{5-(-14)}{5-2}$

$\Rightarrow 4c-7=\frac{5+14}{3}$

$\Rightarrow 12c-21=19$

$\Rightarrow 12c=19+21=40$

$\Rightarrow c=\frac{40}{12}=\frac{10}{3}$

$\therefore c\in (2,5)$

Hence Lagrange's Mean Value theorem is verified.