Question

Verify the  hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = \[\frac{1}{4x - 1},\] 1≤ x ≤ 4 ?

 

Answer 1

The given function is \[f\left( x \right) = \frac{1}{4x - 1}\].

Since for each \[x \in \left[ 1, 4 \right]\] , vthe function attains a unique definite value, \[f\left( x \right)\]  is continuous on \[\left[ 1, 4 \right]\] .

Also,\[f'\left( x \right) = \frac{- 4}{\left( 4x - 1 \right)^2}\] exists for all \[x \in \left[ 1, 4 \right]\]

Thus, both the conditions of Lagrange's mean value theorem are satisfied.
Consequently, there exists some \[c \in \left( 1, 4 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1} = \frac{f\left( 4 \right) - f\left( 1 \right)}{3}\]

Now, 

\[f\left( x \right) = \frac{1}{4x - 1}\]\[\Rightarrow\] \[f'\left( x \right) = \frac{- 4}{\left( 4x -1\right)^2}\] \[f\left( 4 \right) = \frac{1}{15}, f\left( 1 \right) = \frac{1}{3}\]

\[\therefore\] \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1}\]

\[\Rightarrow f'\left( x \right) = \frac{\frac{1}{15} - \frac{1}{3}}{4 - 1} = \frac{- 4}{45}\]

\[ \Rightarrow \frac{- 4}{\left( 4x - 1 \right)^2} = \frac{- 4}{45}\]

\[ \Rightarrow \left( 4x - 1 \right)^2 = 45\]

\[ \Rightarrow 16 x^2 - 8x - 44 = 0\]

\[ \Rightarrow 4 x^2 - 2x - 11 = 0\]

\[ \Rightarrow x = \frac{1}{4}\left( 1 \pm 3\sqrt{5} \right)\]

Thus, \[c = \frac{1}{4}\left( 1 + 3\sqrt{5} \right) \in \left( 1, 4 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1}\]. 

Hence, Lagrange's theorem is verified.