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# 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 $\left[ {1,3} \right]$  Verified
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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.

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.