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:\[f(x) = (x - 1)(x - 2)(x - 3)\]on \[\left[ {0,4} \right]\]

Answer 1

Hint:
Here we have to verify Lagrange’s mean value theorem (LMVT) for the given function \[f(x)\] . Firstly we will check the continuity of the given function in a given interval and then we will check the differentiability of the given function in the given interval. After that to find point \[c\], we will use the formula of LMVT and check if there exists a point \[c\] such that the differentiation of point \[c\] is equal to 0, where \[c\] lies in the given interval.

Formula Used: We will use the formula, \[f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}\].

Complete step by step solution:
Lagrange’s mean value theorem states that if a function is continuous on close interval \[\left[ {a,b} \right]\] and differentiable on open interval \[\left( {a,b} \right)\] then there exists at least one point \[c\] in the given interval such that \[f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}\].
We are given that, \[f(x) = (x - 1)(x - 2)(x - 3)\].
Here the given function \[f(x)\] is continuous on \[\left[ {0,4} \right]\].
Now we know that the given function is a polynomial function thus it is differentiable on \[\left[ {0,4} \right]\].
Now to verify LMVT we have to find \[c \in \left( {0,4} \right)\] such that \[f'(c) = \dfrac{{f(4) - f(0)}}{{4 - 0}}\].
Now, to find \[f(4)\] we take \[x = 4\] in the given function.
Therefore,
\[f(4) = (4 - 1)(4 - 2)(4 - 3)\]
Simplifying the terms, we get
\[ \Rightarrow f\left( 4 \right) = 6\]
And to find \[f(0)\] we take \[x = 0\] in the given function
\[f(0) = (0 - 1)(0 - 2)(0 - 3) = - 6\]
Now we will differentiate the given function.
Therefore, by differentiating given function, we get,
\[f'(x) = 3{x^2} - 12x + 11\]
Substituting \[x = c\] in above equation, we get
\[ \Rightarrow f'(c) = 3{c^2} - 12c + 11\]
Now by putting all these values in the formula of LMVT, \[f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}\], we get
\[f'(c) = \dfrac{{f(4) - f(0)}}{{4 - 0}}\]
\[ \Rightarrow 3{c^2} - 12c + 11 = \dfrac{{6 - ( - 6)}}{4}\]
Now , by solving it further, we get
\[ \Rightarrow 3{c^2} - 12c + 11 = 3\]
Subtracting 3 from both sides, we get
\[ \Rightarrow 3{c^2} - 12c + 8 = 0\]
The obtained equation is a quadratic equation.
Now by solving the above quadratic equation, we get
\[c = \dfrac{{12 \pm \sqrt {144 - 96} }}{6}\]
\[ \Rightarrow c = \dfrac{{12 \pm \sqrt {48} }}{6}\]
By simplifying the equation further, we get
\[ \Rightarrow c = 2 \pm \dfrac{{2\sqrt 3 }}{3}\]
\[ \Rightarrow c = 2 \pm \dfrac{2}{{\sqrt 3 }}\]
Here both the values of \[c\] lie between 0 and 4.
As \[c\] lies between \[\left[ {0,4} \right]\], LMVT is verified.

Note:
Here both values of \[c\] lie in the given interval but in other situations, we have to consider only those values of \[c\] which lie in the given interval and all other values are rejected. At least one value of \[c\] should be obtained otherwise LMVT is not verified.