Question & Answer
QUESTION

Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point ‘\[c\]’ in the indicated interval as stated by the Lagrange's mean value theorem:
 \[f\left( x \right) = 2{x^2} - 3x + 1\] on \[\left[ {1,3} \right]\]

ANSWER Verified Verified
Hint: First of all, we need to check the function \[f\left( x \right)\] satisfies all the conditions of the Lagrange’s mean value theorem. Then find the one point \[c\] in the given interval using Lagrange’s formula. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given \[f\left( x \right) = 2{x^2} - 3x + 1\] on \[\left[ {1,3} \right]\]
Lagrange’s Mean Value Theorem (LMVT) states that if a function \[f\left( x \right)\] is continuous on a closed interval \[\left[ {a,b} \right]\] and differentiable on the open interval \[\left( {a,b} \right)\], then there is at least one point \[x = c\]on this interval, such that \[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\].

We know that a polynomial function is continuous everywhere and also differential.

So, \[f\left( x \right)\]being a polynomial is continuous on \[\left[ {1,3} \right]\] and differentiable on \[\left( {1,3} \right)\]

So, there must exist at least one real number \[c \in \left( {1,3} \right)\]
\[ \Rightarrow f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}} = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}\]

We have \[f\left( x \right) = 2{x^2} - 3x + 1\]
So \[f\left( 1 \right) = 2{\left( 1 \right)^2} - 3\left( 1 \right) + 1 = 2 - 3 + 1 = 0\]
     \[f\left( 3 \right) = 2{\left( 3 \right)^2} - 3\left( 3 \right) + 1 = 18 - 9 + 1 = 10\]

And \[f'\left( x \right) = \dfrac{{d\left( {2{x^2} - 3x + 1} \right)}}{{dx}} = 4x - 3\]
So \[f'\left( c \right) = 4c - 3\]

As \[f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}\], we have
\[
   \Rightarrow 4c - 3 = \dfrac{{10 - 0}}{2} \\
   \Rightarrow 4c - 3 = 5 \\
   \Rightarrow 4c = 5 + 3 \\
   \Rightarrow 4c = 8 \\
   \Rightarrow c = 2 \\
  \therefore c \in \left( {1,3} \right) \\
\]

Note: This theorem (also known as First mean value Theorem) allows to express the increment of a function on an interval through the value of derivative at an intermediate point of segment.