Question

Find $c$ of Lagrange's mean value theorem $f\left(x\right)=x(x-1)(x-2)$ where $x\in (0,\frac{1}{2})$

Answer 1

We have $f\left(x\right)=x(x-1)(x-2)=x^3-3x^2+2x$

$\left(i\right)f\left(x\right)$ is a polynomial function and is continuous in $[0,\frac{1}{2}]$

$\left(ii\right)f^{\prime }\left(x\right)=3x^2-6x+2$ which exists and hence is differentiable in $(0,\frac{1}{2})$.

Now $f^{\prime }\left(c\right)=3c^2-6c+2$

$f\left(0\right)=0,f\left(\frac{1}{2}\right)=\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)=\frac{1}{2}\times \frac{1}{2}\times \frac{-3}{2}=\frac{-3}{8}$

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

$\frac{\frac{3}{8}-0}{\frac{1}{2}-0}=3c^2-6c+2$

$\Rightarrow \frac{3}{8}\times \frac{1}{2}=3c^2-6c+2$

$\Rightarrow 12c^2-24c+5=0$

$\Rightarrow c=\frac{24\pm \sqrt{{(-24)}^2-4\times 12\times 5}}{2\times 12}$

$\Rightarrow c=1-\frac{\sqrt{21}}{6}=0.236$(approx.)

And $c=1+\frac{\sqrt{21}}{6}>\frac{1}{2}$ is not acceptable.