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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(x) = x3 − 5x2 − 3x on [1, 3] ?
Solution
We have,
\[f\left( x \right) = x^3 - 5 x^2 - 3x\]
Since polynomial function is everywhere continuous and differentiable
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 1, 3 \right]\] and differentiable on \[\left( 1, 3 \right)\]
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some
∴ \[f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]
\[\Rightarrow 3 x^2 - 10x - 3 = \frac{- 20}{2}\]
\[ \Rightarrow 3 x^2 - 10x + 7 = 0\]
\[ \Rightarrow x = 1, \frac{7}{3}\]
Thus, \[c = \frac{7}{3} \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\] .
Hence, Lagrange's theorem is verified.
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