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प्रश्न
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) = x(x + 4)2 on [0, 4] ?
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उत्तर
We have,
\[f\left( x \right) = x \left( x + 4 \right)^2 = x\left( x^2 + 16 + 8x \right) = x^3 + 8 x^2 + 16x\]
Since \[f\left( x \right)\] is a polynomial function which is everywhere continuous and differentiable.
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 0, 4 \right]\] and derivable on \[\left( 0, 4 \right)\]
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some
\[\Rightarrow 3 x^2 + 16x + 16 = \frac{256}{4}\]
\[ \Rightarrow 3 x^2 + 16x - 48 = 0\]
\[ \Rightarrow x = - \frac{4}{3}\left( 2 + \sqrt{13} \right), \frac{4}{3}\left( \sqrt{13} - 2 \right)\]
Thus,\[c = \frac{- 8 + 4\sqrt{13}}{3} \in \left( 0, 4 \right)\] 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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