Question

Verify Langrange's mean value theorem for the function:
$f\left(x\right)=x(1-\log x)$ and find the value of '$c$' in the interval [1, 2].

Answer 1

$f\left(x\right)=x(1-\log x)$

The function $f\left(x\right)$ is continuous in $[1,2]$ because $x$ is continuous every where while $\log x$ is continuous when $xϵ[1,\infty ]$

The function $f\left(x\right)$ is also differential in $(1,2)$

Therefore there exist $c^{\prime }$ , such that $c$ belongs to $(1,2)$

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

$f^{\prime }\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

$f^{\prime }\left(c\right)=\frac{(2-2\log 2)-(1-\log 1)}{2-1}$

$f^{\prime }\left(c\right)=\frac{1-2\log 2}{2-1}$

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

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

$-\log c=1-2\log 2$

$\log c=2\log 2-1$

$\log c=\log 4-1$

$\log 4-\log c=1$

$\log \frac{4}{c}=1$

$\frac{4}{c}=e^1$

$c=\frac{4}{e^1}$

$c=1.472$

Thus the value of $c$ belongs to $(1,2)$