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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 theore f(x) = (x − 1)(x − 2)(x − 3) on [0, 4] ?
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Solution
We have,
\[f\left( x \right) = \left( x - 1 \right)\left( x - 2 \right)\left( x - 3 \right)\] which can be rewritten as
Therefore,
So, there must exist at least one real number \[c \in \left( 0, 4 \right)\] such that
\[\Rightarrow 3 x^2 - 12x + 11 = \frac{6 + 6}{4}\]
\[ \Rightarrow 3 x^2 - 12x + 8 = 0\]
\[ \Rightarrow x = 2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}\]
Thus,\[c = 2 \pm \frac{2}{\sqrt{3}} \in \left( 0, 4 \right)\] such that\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\] .
Hence, Lagrange's theorem is verified.
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