Question

Find $c$ using Langrange's mean value theorem.
$F\left(x\right)=\frac{x^2-4x}{x+2}[0,4]$

Answer 1

Find C using language's main value.

$F\left(x\right)=\frac{x^2-4x}{x+2}[0,4]$

$\to$ condition for mean value theorem

1. f(x) is continuous at (a,b)

2. f(x) is derivable at (a,b)

If both condition satisfied then these exists

some c in (a,b) such that

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$\to f\left(x\right)=\frac{x^2-4x}{x+2}$ a = 0 , b = 4

$f^{\prime }\left(x\right)=\frac{(x+2)[2x-4]-(x^2-4x)\left(1\right)}{(x+2{)}^2}$

$=\frac{2x^2-4x+4x-8-x^2+4x}{(x+2{)}^2}$

$f^{\prime }\left(x\right)=\frac{x^2+4x-8}{(x+2{)}^2}$

$f^{\prime }\left(c\right)=\frac{c^2+4c-8}{(c+2{)}^2}$

$\to f\left(b\right)=\frac{(a{)}^2-4(h)}{4+2}=0$

$\to f\left(a\right)=\frac{(0{)}^2-4(0)}{0+2}=0$

$\to f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$\therefore \frac{c^2+4c-8}{(c+2{)}^2}=\frac{0}{4}$

$\therefore c^2+4c-8=0$ & $(c+2{)}^2\ne 0$

$\therefore$ root of the equation $\alpha ,\beta =\frac{-4\pm \sqrt{16-4\times 8}}{2}$

$\alpha ,\beta =\frac{-4\pm \sqrt{-16}}{2}=\frac{-4\pm \sqrt{16i2}}{2}=\frac{-4\pm 4i}{2}$