Question

If a function is given as $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ , $x\in \left[ 0,4 \right]$ , Then find $'c'$ if LMVT can be applied.

Answer 1

Hint: For solving this question first we will see the result of LMVT that if a function $f\left( x \right)$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ then there exists a point $x=c$ in $\left( a,b \right)$ such that ${f}'\left( c \right)=0$ . After that, we will differentiate the given function with respect to $x$ and solve for the value of $x$ for which ${f}'\left( x \right)=0$ and give the final answer for this question.

Complete step-by-step solution -
Given:
We have a function $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ , $x\in \left[ 0,4 \right]$ , and we have to find the value of $'c'$ if LMVT can be applied.
Now, before we proceed we should know what LMVT is.
Lagrange’s Mean Value Theorem (LMVT):
If a function $f\left( x \right)$ is defined on the closed interval $\left[ a,b \right]$ satisfying the following conditions:
1. The function $f\left( x \right)$ is continuous on the closed interval $\left[ a,b \right]$
2. The function $f\left( x \right)$ is differentiable on the open interval $\left( a,b \right)$
Then there exists a value $x=c$ in such a way that ${f}'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{b-a}$ . This is also known as the first mean value theorem or Lagrange’s mean value theorem (LMVT).
Now, a special case of Lagrange’s mean value theorem is Rolle’s Theorem which states that, if a function $f\left( x \right)$ is defined on the closed interval $\left[ a,b \right]$ satisfying the following conditions:
1. The function $f\left( x \right)$ is continuous on the closed interval $\left[ a,b \right]$
2. The function $f\left( x \right)$ is differentiable on the open interval $\left( a,b \right)$
3. Now if $f\left( a \right)=f\left( b \right)$ , then there exists at least one value of $x$ , let us assume this value to be $c$ , which lies between $a$ and $b$ i.e. $\left( aPrecisely, we conclude that according to LMVT if a function $f\left( x \right)$ is continuous on the closed interval $\left[ a,b \right]$ and differentiable on the open interval $\left( a,b \right)$ then there exists a point $x=c$ in $\left( a,b \right)$ such that ${f}'\left( c \right)=0$ .
Now, we come back to our question in which we have a function $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ and we have to find the value of $c$ for which LMVT can be applied.
Now, it is evident that the value of the given function $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ at $x=1,2,3$ will be zero. So, from LMVT we can say that there must be two values of the $x$ for which ${f}'\left( x \right)=0$ .
Now, we will write $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ in some other form and differentiate it with respect to $x$ . Then,
$\begin{align}
  & f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\
 & \Rightarrow f\left( x \right)=\left( {{x}^{2}}-3x+2 \right)\left( x-3 \right) \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-3{{x}^{2}}+2x-3{{x}^{2}}+9x-6 \\
 & \Rightarrow f\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6 \\
 & \Rightarrow \dfrac{d\left( f\left( x \right) \right)}{dx}=\dfrac{d\left( {{x}^{3}}-6{{x}^{2}}+11x-6 \right)}{dx} \\
 & \Rightarrow {f}'\left( x \right)=3{{x}^{2}}-12x+11 \\
\end{align}$
Now, we will find the value of $x$ for which value of ${f}'\left( x \right)=3{{x}^{2}}-12x+11=0$ . Then,
$3{{x}^{2}}-12x+11=0$
Now, we know that if $a{{x}^{2}}+bx+c=0$ then the value of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . So, the value of $a=3$ , $b=-12$ and $c=11$ . Then,
\[\begin{align}
  & 3{{x}^{2}}-12x+11=0 \\
 & \Rightarrow x=\dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\times 3\times 11}}{2\times 3} \\
 & \Rightarrow x=\dfrac{12\pm \sqrt{144-132}}{6} \\
 & \Rightarrow x=\dfrac{12\pm \sqrt{12}}{6} \\
\end{align}\]

\[\begin{align}
  & \Rightarrow x=\dfrac{12\pm 2\sqrt{3}}{6} \\
 & \Rightarrow x=\dfrac{12}{6}\pm \dfrac{2\sqrt{3}}{6} \\
 & \Rightarrow x=2\pm \dfrac{1}{\sqrt{3}} \\
\end{align}\]
Now, from the above result, we conclude that value of ${f}'\left( x \right)=3{{x}^{2}}-12x+11=0$ for the following two values of $x$ :
1. $x=2+\dfrac{1}{\sqrt{3}}\approx 2.577$ which is between 2 and 3
2. $x=2-\dfrac{1}{\sqrt{3}}\approx 1.422$ which is between 1 and 2
Now, we conclude that for the function $f\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ , $x\in \left[ 0,4 \right]$ we have two values of $c=1.422,2.577$ for which ${f}'\left( x \right)=0$ and LMVT can be applied.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer. After that, we should apply the concept of LMVT and the solution of quadratic equation $a{{x}^{2}}+bx+c=0$ with suitable values and proceed stepwise. Moreover, we should avoid calculation mistakes while solving to get the correct answer.