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 − 5x² − 3x on [1, 3] ?

Answer 1

We have,

\[f\left( x \right) = x^3 - 5 x^2 - 3x\]

Since polynomial function is everywhere continuous and differentiable
Therefore, \[f\left( x \right)\] is continuous on \[\left[ 1, 3 \right]\] and differentiable on \[\left( 1, 3 \right)\]

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some 

\[c \in \left( 1, 3 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1} = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]

Now, \[f\left( x \right) = x^3 - 5 x^2 - 3x\]

\[f'\left( x \right) = 3 x^2 - 10x - 3\],\[f\left( 3 \right) = - 27\],\[f\left( 1 \right) = - 7\]

∴  \[f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]

\[\Rightarrow 3 x^2 - 10x - 3 = \frac{- 20}{2}\]

\[ \Rightarrow 3 x^2 - 10x + 7 = 0\]

\[ \Rightarrow x = 1, \frac{7}{3}\]

Thus, \[c = \frac{7}{3} \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\] .

Hence, Lagrange's theorem is verified.