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) = tan⁻¹ x on [0, 1] ?

Answer 1

 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.