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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 − 2x2 − x + 3 on [0, 1] ?
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Solution
We have,
\[f\left( x \right) = x^3 - 2 x^2 - x + 3 = 0\]
Since a polynomial function is everywhere continuous and differentiable.
Therefore,
\[f\left( x \right)\] is continuous on \[\left[ 0, 1 \right]\] and differentiable on \[\left( 0, 1 \right)\]
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number \[c \in \left( 0, 1 \right)\] such that \[f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}\]
Now, \[f\left( x \right) = x^3 - 2 x^2 - x + 3 = 0\]
\[\Rightarrow f'\left( x \right) = 3 x^2 - 4x - 1\] ,
\[\Rightarrow 3 x^2 - 4x - 1 = \frac{1 - 3}{1}\]
\[ \Rightarrow 3 x^2 - 4x - 1 + 2 = 0\]
\[ \Rightarrow 3 x^2 - 4x + 1 = 0\]
\[ \Rightarrow 3 x^2 - 3x - x + 1 = 0\]
\[ \Rightarrow \left( 3x - 1 \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow x = \frac{1}{3}, 1\]
Thus,
\[c = \frac{1}{3} \in \left( 0, 1 \right)\] 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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