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Question: Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each c...

Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point ‘cc’ in the indicated interval as stated by the Lagrange's mean value theorem:
f(x)=2x2−3x+1f\left( x \right) = 2{x^2} - 3x + 1 on [1,3]\left[ {1,3} \right]

Explanation

Solution

Hint: First of all, we need to check the function f(x)f\left( x \right) satisfies all the conditions of the Lagrange’s mean value theorem. Then find the one point cc 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(x)=2x2−3x+1f\left( x \right) = 2{x^2} - 3x + 1 on [1,3]\left[ {1,3} \right]
Lagrange’s Mean Value Theorem (LMVT) states that if a function f(x)f\left( x \right) is continuous on a closed interval [a,b]\left[ {a,b} \right] and differentiable on the open interval (a,b)\left( {a,b} \right), then there is at least one point x=cx = con this interval, such that f′(c)=f(b)−f(a)b−af'\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(x)f\left( x \right)being a polynomial is continuous on [1,3]\left[ {1,3} \right] and differentiable on (1,3)\left( {1,3} \right)

So, there must exist at least one real number c∈(1,3)c \in \left( {1,3} \right)
⇒f′(c)=f(3)−f(1)3−1=f(3)−f(1)2\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(x)=2x2−3x+1f\left( x \right) = 2{x^2} - 3x + 1
So f(1)=2(1)2−3(1)+1=2−3+1=0f\left( 1 \right) = 2{\left( 1 \right)^2} - 3\left( 1 \right) + 1 = 2 - 3 + 1 = 0
f(3)=2(3)2−3(3)+1=18−9+1=10f\left( 3 \right) = 2{\left( 3 \right)^2} - 3\left( 3 \right) + 1 = 18 - 9 + 1 = 10

And f′(x)=d(2x2−3x+1)dx=4x−3f'\left( x \right) = \dfrac{{d\left( {2{x^2} - 3x + 1} \right)}}{{dx}} = 4x - 3
So f′(c)=4c−3f'\left( c \right) = 4c - 3

As f′(c)=f(3)−f(1)2f'\left( c \right) = \dfrac{{f\left( 3 \right) - f\left( 1 \right)}}{2}, we have

⇒4c−3=10−02 ⇒4c−3=5 ⇒4c=5+3 ⇒4c=8 ⇒c=2 ∴c∈(1,3)  \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.