$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$
Here, f(x) will exist,
if
$\text{x}^2-4\geq0$
$\Rightarrow\text{x}\leq-2\text{ or }\text{x}\geq2$
Since, for each $\text{x}\in2,4,$ the function f(x) attains a unique definite value.
So, f(x) is continuous on 2, 4
Also,
$\text{f}'(\text{x})=\frac{1}{2\sqrt{\text{x}^2-4}}(2\text{x})=\frac{\text{x}}{\sqrt{\text{x}^2-4}}$
Exists for all $\text{x}\in2,4$
So, f(x) is differentiable on 2, 4.
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists some $\text{c}\in2,4$ such that
$\text{f}'(\text{x})=\frac{1}{2\sqrt{\text{x}^2-4}}(2\text{x})=\frac{\text{x}}{\sqrt{\text{x}^2-4}}$
Now,
$\text{f}(\text{x})=\sqrt{\text{x}^2-4}$
$\text{f}'(\text{x})=\frac{1}{\sqrt{\text{x}^2-4}},\text{f}(4)=2\sqrt3,\text{f}(2)=0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(2)}{4-2}$
$\Rightarrow\frac{\text{x}}{\sqrt{\text{x}^2-4}}=\frac{2\sqrt3}{2}$
$\Rightarrow\frac{\text{x}}{\sqrt{\text{x}^2-4}}=\sqrt3$
$\Rightarrow\frac{\text{x}^2}{\text{x}^2-4}=3$
$\Rightarrow\text{x}^2=3\text{x}^2-12$
$\Rightarrow\text{x}^2=6$
$\Rightarrow\text{x}=\pm\sqrt6$
Thus, $\text{c}=\sqrt6\in(2,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(2)}{4-2}$
Hence, Lagrange's theorem is verified.
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| $\text{x}_\text{i}$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $\text{p}_\text{i}$ | $\frac{1}{6}$ | $\frac{5}{18}$ | $\frac{2}{9}$ | $\frac{1}{6}$ | $\frac{1}{9}$ | $\frac{1}{18}$ |