Question

Verify Lagrange's Mean Value Theorem (LMVT) for following functions on indicated intervals. Also, find a point c in the indicated interval that satisfy LMVT.
i) $f\left(x\right)=x(x-2)$ on $[1,3]$
ii) $f\left(x\right)=x(x-1)(x-3)$ on $[0,1]$

Answer 1

LMVT states that,
If $g:[a,b]\to R$ is a continuous function on $[a,b]$ and differentiable on $(a,b)$, then there exists some $c$ in $(a,b)$ such that $g^{\prime }\left(c\right)=\frac{g\left(b\right)-g\left(a\right)}{b-a}$

i) $f\left(x\right)=x(x-2)$ on $[1,3]$
We know that, a polynomial function is everywhere continuous and differentiable. So, $f\left(x\right)$ being a polynomial, is continuous on $[1,3]$ and differentiable on $(1,3)$.


Thus, $f\left(x\right)$ satisfies both the conditions of LMVT on $[1,3]$.
So, there must exist at least one real number $c\in (1,3)$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(3\right)-f\left(1\right)}{3-1}$
$f^{\prime }\left(x\right)=2x-2,f^{\prime }\left(c\right)=2c-2,f\left(3\right)=9-6=3$ and $f\left(1\right)=1-2=-1$
$f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}\Rightarrow 2c-2=\frac{3-(-1)}{3-1}\Rightarrow 2c-2=2\Rightarrow c=2$
Thus, $c=2\in (1,3)$ such that $f^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$
$\Rightarrow$ LMVT is verified.

ii) $f\left(x\right)=x(x-1)(x-3)$ on $[0,1]$
Since, a polynomial function is everywhere continuous and differentiable. So, $f\left(x\right)$ being a polynomial, is continuous on $[0,1]$ and differentiable on $(0,1)$. Thus, $f\left(x\right)$ satisfies both the conditions of LMVT on $[0,1]$.
So, there must exist at least one real number $c\in (0,1)$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
$f^{\prime }\left(x\right)=3x^2-8x+3,f^{\prime }\left(c\right)=3c^2-8c+3,f\left(1\right)=0$ and $f\left(0\right)=0$.
$\Rightarrow 3c^2-8c+3=\frac{0-0}{1-0}=0\Rightarrow c=\frac{8\pm \sqrt{64-36}}{6}=\frac{8\pm \sqrt{28}}{6}$
$\Rightarrow c=\frac{4-\sqrt{7}}{3}=0.451\in (0,1)$ such that $f^{\prime }\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
$\Rightarrow$ LMVT is verified.