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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 + 1 X on [ 1 , 3 ] - CBSE (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 $f\left( x \right) = x + \frac{1}{x} \text { on }[1, 3]$ ?

#### Solution

We have,

$f\left( x \right) = x + \frac{1}{x} = \frac{x^2 + 1}{x}$

Clearly,

$f\left( x \right)$ is continuous on $\left[ 1, 3 \right]$ and derivable on $\left( 1, 3 \right)$

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some  $c \in \left( 1, 3 \right)$ such that $f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1} = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}$

Now,

$f\left( x \right) = \frac{x^2 + 1}{x}$

$f'\left( x \right) = \frac{x^2 - 1}{x^2}$

$f\left( 1 \right) = 2$,
$f\left( 3 \right) = \frac{10}{3}$
∴$f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}$

$\Rightarrow \frac{x^2 - 1}{x^2} = \frac{4}{6}$

$\Rightarrow \frac{x^2 - 1}{x^2} = \frac{2}{3}$

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

$\Rightarrow x = \pm \sqrt{3}$

Thus,

$c = \sqrt{3} \in \left( 1, 3 \right)$ such that
$f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}$ .
Hence, Lagrange's theorem is verified.
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#### APPEARS IN

Solution for 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 F ( X ) = X + 1 X on [ 1 , 3 ] concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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