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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) = Tan−1 X on [0, 1] ? - 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 theore  f(x) = tan1 x on [0, 1] ?

Sum
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Solution

 We have,

\[f\left( x \right) = \tan^{- 1} x\]

Clearly.  \[f\left( x \right)\] is continuous on \[\left[ 0, 1 \right]\] and derivable on \[\left( 0, 1 \right)\]

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some \[c \in \left( - 3, 4 \right)\] such that \[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0} = \frac{f\left( 1 \right) - f\left( 0 \right)}{1}\]
Now, 
\[f\left( x \right) = \tan^{- 1} x\]\[f'\left( x \right) = \frac{1}{1 + x^2}\] ,\[f\left( 1 \right) = \frac{\pi}{4}\] ,\[f\left( 0 \right) = 0\]
∴  \[f'\left( x \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\]

\[\Rightarrow \frac{1}{1 + x^2} = \frac{\pi}{4} - 0\]

\[ \Rightarrow \left( \frac{4}{\pi} - 1 \right) = x^2 \]

\[ \Rightarrow x = \pm \sqrt{\frac{4 - \pi}{\pi}}\]

Thus, \[c = \sqrt{\frac{4 - \pi}{\pi}} \in \left( 0, 1 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\] .
Hence, Lagrange's theorem is verified.

 

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Chapter 15: Mean Value Theorems - Exercise 15.2 [Page 17]

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RD Sharma Mathematics [English] Class 12
Chapter 15 Mean Value Theorems
Exercise 15.2 | Q 1.1 | Page 17

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