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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 Mean Valu F(X) = X2 − 1 on [2, 3] ? - CBSE (Science) Class 12 - Mathematics

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(x) = x2 − 1 on [2, 3] ?

#### Solution

We have

$f\left( x \right) = x^2 - 1$

Since a polynomial function is everywhere continuous and differentiable, $f\left( x \right)$ is continuous on $\left[ 2, 3 \right]$ and differentiable on $\left( 2, 3 \right)$.

Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ $c \in \left( 2, 3 \right)$ such that $f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}$

Now,

$f\left( x \right) = x^2 - 1$
$\Rightarrow f'\left( x \right) = 2x$ ,
$f\left( 3 \right) = \left( 3 \right)^2 - 1 = 8$ ,
$f\left( 2 \right) = \left( 2 \right)^2 - 1 = 3$
$f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}$

$\Rightarrow 2x = \frac{8 - 3}{1}$

$\Rightarrow x = \frac{5}{2}$

Thus,

$c = \frac{5}{2} \in \left( 2, 3 \right)$ such that $f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 2 \right)}{3 - 2}$.

Hence, Lagrange's theorem is verified.

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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 Mean Valu F(X) = X2 − 1 on [2, 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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