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Question
Verify the hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = \[\frac{1}{4x - 1},\] 1≤ x ≤ 4 ?
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Solution
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 \[x \in \left[ 1, 4 \right]\]
Consequently, there exists some \[c \in \left( 1, 4 \right)\] such that
\[\Rightarrow f'\left( x \right) = \frac{\frac{1}{15} - \frac{1}{3}}{4 - 1} = \frac{- 4}{45}\]
\[ \Rightarrow \frac{- 4}{\left( 4x - 1 \right)^2} = \frac{- 4}{45}\]
\[ \Rightarrow \left( 4x - 1 \right)^2 = 45\]
\[ \Rightarrow 16 x^2 - 8x - 44 = 0\]
\[ \Rightarrow 4 x^2 - 2x - 11 = 0\]
\[ \Rightarrow x = \frac{1}{4}\left( 1 \pm 3\sqrt{5} \right)\]
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.
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