Question

Verify Lagrange's mean value theorem for

f (x) = \sqrt{x^{2} - 4}

in the interval

[2, 4]

.

Accepted Answer

A function $f (x)$ satisfies Lagrange's mean value theorem in $[a, b],$ if

$(i) f (x)$ is continuous in $[a, b],$

$(i i) f (x)$ is differentiable in $(a, b),$ then, there exist atleast one $c \in (a, b)$ such that $f^{'} (c) = \frac{f (b) - f (a)}{b - a} .$

Given $f (x) = \sqrt{x^{2} - 4}, x \in [2, 4] .$

Clearly $f (x)$ is defined for all $x \in [2, 4] .$

We know that, the polynomial function is continuous and differentiable everywhere, hence

$(i)$ $f (x)$ is continuous on $[2, 4],$

$(i i)$ $f (x)$ is differentiable on $(2, 4)$ ,

Hence, the function satisfies all conditions of Lagrange's mean value theorem, then, their exist atleast once value $c \in (2, 4)$ such that

$f^{'} (c) = \frac{f (4) - f (2)}{4 - 2} .$

Now $f (2) = \sqrt{2^{2} - 4} = 0$ and $f (4) = \sqrt{4^{2} - 4} = \sqrt{12} = 2 \sqrt{3} .$

And $f (x) = \sqrt{x^{2} - 4} .$

Using chain rule for finding the derivative of a composite function i.e. $\frac{d}{d x} h (g (x)) = h^{'} (g (x)) \times (g^{'} (x)),$

$\Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x^{2} - 4}} (2 x) = \frac{x}{\sqrt{x^{2} - 4}} .$

Taking $f^{'} (c) = \frac{f (4) - f (2)}{4 - 2}$

$\Rightarrow \frac{c}{\sqrt{c^{2} - 4}} = \frac{2 \sqrt{3} - 0}{2}$

$\Rightarrow c = \sqrt{3} \times \sqrt{c^{2} - 4}$

$\Rightarrow c^{2} = 3 (c^{2} - 4)$

$\Rightarrow c^{2} = 3 c^{2} - 12$

$\Rightarrow 2 c^{2} = 12$

$\Rightarrow c = \pm \sqrt{6}, \sqrt{6} \in (2, 4) .$

Hence, the Lagrange's mean value theorem is satisfied for $f (x) = \sqrt{x^{2} - 4}$ at $x = \sqrt{6}$ in the interval $[2, 4] .$

Embibe Experts Solutions for Exercise 5: Assignment

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 5: Assignment

Questions from Embibe Experts Solutions for Exercise 5: Assignment with Hints & Solutions

Learn and Prepare for any exam you want !