Question

Verify Lagrange's Mean Value Theorem for the function $f\left(x\right)=\sqrt{x^2-x}$ in the interval $[1,4].$

Answer 1

Given, $f\left(x\right)=\sqrt{x^2-x},x\in [1,4]$
Since $x^2-x$ is continuous on $R$
and $\sqrt{x^2-x}$ is exists in $[1,4]$
$\Rightarrow f\left(x\right)$ is continuous in $[1,4].$

Differentiating the given function w.r.t. $x$
$f^{\prime }\left(x\right)=\frac{1}{2}(x^2-x{)}^{-1/2}.(2x-1)=\frac{2x-1}{2\sqrt{x^2-x}}$
which exists $\forall x\in R$.

$\therefore f\left(x\right)$ is differentiable in $(1,4)$.

Thus, both the conditions of Lagrange's mean value theorem is satisfied therefore, $\exists c$ in $(1,4).$
Such that $f^{\prime }\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$

$\frac{2c-1}{2\sqrt{c^2-c}}=\frac{\sqrt{12}}{3}$

$3(2c-1)=2\sqrt{c^2-c}.\sqrt{12}$

$9(4c^2-4c+1)=48(c^2-c)$

$3(4c^2-4c+1)=16(c^2-c)$

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

$\Rightarrow 4c^2-4c-3=0$

$\Rightarrow (2c-3)(2c+1)=0$

$c=\frac{3}{2},\frac{-1}{2}$

Thus, $\exists$ (there exist) $c=\frac{3}{2}\in (1,4)$

Such that $f^{\prime }\left(\frac{3}{2}\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$

Hence, Lagrange's mean value theorem is verified and $c=\frac{3}{2}$.