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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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.

  Is there an error in this question or solution?
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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