Question

Verify Lagrange's mean value theorem for the following function on the indicated interval. In each case find a point ${}^{\prime }{c}^{\prime }$ in the indicated interval as stated by the Lagrange's mean value theorem:
$f\left(x\right)=x(x+4{)}^2$ on $[0,4]$

Answer 1

$f\left(x\right)=x{(x+4)}^2=x(x^2+8x+16)=x^3+8x^2+16x$ on $[0,4]$

$x^3+8x^2+16x$ has unique value for all $x$ between $0$ and $4$.

$\therefore f\left(x\right)$ is continuous in $[0,4]$

$f\left(x\right)=x^3+8x^2+16x$

Differentiating with respect to $x:$

$f^{\prime }\left(x\right)=3x^2+16x+16$

$\therefore f\left(x\right)$ is differentiable in $(0,4)$

So both the necessary conditions of Lagrange’s mean value theorem is satisfied.

$\therefore$ there exists a point $c\in (0,4)$ such that:

$f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

$\Rightarrow f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4}$

We have $f^{\prime }\left(x\right)=3x^2+16x+16$

$f^{\prime }\left(c\right)=3c^2+16c+16$

For $f\left(4\right)=3\times 4^2+16\times 4+16=48+64+16=128$

$f\left(0\right)=0+0+16=16$

$f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4}$

$\Rightarrow 3c^2+16c+16=\frac{128-16}{4}$

$\Rightarrow 3c^2+16c+16=\frac{112}{4}=28$

$\Rightarrow 3c^2+16c+16-28=0$

$\Rightarrow 3c^2+16c-12=0$

$\Rightarrow c=\frac{-16\pm \sqrt{256+144}}{6}=\frac{-16\pm 20}{6}$

$\therefore c=\frac{-16+20}{6}=\frac{4}{6}=\frac{2}{3}\in (0,4)$

$\therefore c=\frac{-16-20}{6}=\frac{-36}{6}=-6\notin (0,4)$

Hence $c=\frac{2}{3}$

Hence, Lagrange’s mean value theorem is verified.