PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the Lagrange'S F ( X ) = √ X 2 − 4 on [ 2 , 4 ] ? - PUC Karnataka Science Class 12 - Mathematics

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ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem $f\left( x \right) = \sqrt{x^2 - 4} \text { on }[2, 4]$ ?

#### Solution

We have,

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

Here,

$f\left( x \right)$ will exist,
if $x^2 - 4 \geq 0$

$\Rightarrow x \leq - 2 \text { or } x \geq 2$

Since for each  $x \in \left[ 2, 4 \right]$  the function

$f\left( x \right)$ attains a unique definite value.
So,
$f\left( x \right)$ is continuous on $\left[ 2, 4 \right]$
Also,
$f'\left( x \right) = \frac{1}{2\sqrt{x^2 - 4}}\left( 2x \right) = \frac{x}{\sqrt{x^2 - 4}}$ exists for all $x \in \left( 2, 4 \right)$
So,
$f\left( x \right)$ is differentiable on $\left( 2, 4 \right)$ .
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some
$c \in \left( 2, 4 \right)$ such that
$f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 2 \right)}{4 - 2} = \frac{f\left( 4 \right) - f\left( 2 \right)}{2}$
Now,
$f\left( x \right) = \sqrt{x^2 - 4}$
$f'\left( x \right) = \frac{x}{\sqrt{x^2 - 4}}$ ,
$f\left( 4 \right) = 2\sqrt{3}$ ,
$f\left( 2 \right) = 0$
∴ $f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 2 \right)}{4 - 2}$

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

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

$\Rightarrow \frac{x^2}{x^2 - 4} = 3$

$\Rightarrow x^2 = 3 x^2 - 12$

$\Rightarrow x^2 = 6$

$\Rightarrow x = \pm \sqrt{6}$

Thus,

$c = \sqrt{6} \in \left( 2, 4 \right)$ such that

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

Hence, Lagrange's theorem is verified.

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#### APPEARS IN

Solution Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the Lagrange'S F ( X ) = √ X 2 − 4 on [ 2 , 4 ] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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