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) = x² + x − 1 on [0, 4] ?

Answer 1

We have,

\[f\left( x \right) = x^2 + x - 1\]

Since polynomial function is everywhere continuous and differentiable.
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 0, 4 \right]\] and differentiable 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, \[f\left( x \right) = x^2 + x - 1\]

\[f'\left( x \right) = 2x + 1\],\[f\left( 4 \right) = 19\],\[f\left( 0 \right) = - 1\]

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

\[\Rightarrow 2x + 1 = \frac{20}{4}\]

\[ \Rightarrow 2x + 1 = 5\]

\[ \Rightarrow 2x = 4 \]

\[ \Rightarrow x = 2\]

Thus, \[c = 2 \in \left( 0, 4 \right)\] such that \[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}\].

Hence, Lagrange's theorem is verified.