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

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.