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) = x(x + 4)² on [0, 4] ?

Answer 1

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 

\[c \in \left( 0, 4 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0} = \frac{f\left( 4 \right) - f\left( 0 \right)}{4}\]

Now, \[f\left( x \right) = x^3 + 8 x^2 + 16x\]

\[f'\left( x \right) = 3 x^2 + 16x + 16\],\[f\left( 4 \right) = 64 + 128 + 64 = 256\] ,\[f\left( 0 \right) = 0\]

∴  \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\]

\[\Rightarrow 3 x^2 + 16x + 16 = \frac{256}{4}\]

\[ \Rightarrow 3 x^2 + 16x - 48 = 0\]

\[ \Rightarrow x = - \frac{4}{3}\left( 2 + \sqrt{13} \right), \frac{4}{3}\left( \sqrt{13} - 2 \right)\]

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.