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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

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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.

  Is there an error in this question or solution?
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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