Question

What are the values of \[c\] for which Rolle`s Theorem for the function $f\left( x \right) = {x^3} - 3{x^2} + 2x$ in the interval $\left[ {0,2} \right]$is verified?
A. \[c = \pm 1\]
B. \[c = 1 \pm \dfrac{1}{{\sqrt 3 }}\]
C. \[c = \pm 2\]
D. None of these

Answer 1

Hint: Rolle`s Theorem must satisfy all the three conditions and if $f\left( x \right)$ is a polynomial function then the function is continuous in the interval.
First of all, we should know the conditions of Rolle`s theorem
Conditions for Rolle`s theorem is
(1) $f\left( x \right)$ is a continuous at $\left[ {a,b} \right]$
(2) $f\left( x \right)$is derivable at $\left( {a,b} \right)$
(3) $f\left( a \right) = f\left( b \right)$
If all the three conditions satisfied then there exist some \[c\] in $f\left( a \right) = f\left( b \right)$ such that $f'\left( c \right) = 0$
Here we can clearly observe that
(a) $f\left( x \right)$ is a polynomial, so it is continuous in the interval $\left[ {0,2} \right]$
(b) $f'\left( x \right) = 3{x^2} - 6x + 2$ exists for all $x \in \left( {0,2} \right)$
So, $f\left( x \right)$ is differentiable for all $x \in \left( {0,2} \right)$ and
(c) $f\left( 0 \right) = f\left( 2 \right)$ since,
$
  f\left( 0 \right) = {\left( 0 \right)^3} - 3{\left( 0 \right)^2} + 2\left( 0 \right) = 0 \\
  f\left( 2 \right) = {\left( 2 \right)^3} - 3{\left( 2 \right)^2} + 2\left( 2 \right) \\
  {\text{ = }}8 - 3 \times 4 + 4 = 0 \\
$
Thus, all the three conditions of Rolle`s theorem are satisfied.
So, there must be exist $c \in \left[ {0,2} \right]$ such that $f'\left( c \right) = 0$
$f'\left( c \right) = 3{\left( c \right)^2} - 6c + 2 = 0$
solving the equation $3{\left( c \right)^2} - 6c + 2 = 0$ we get
$
   \Rightarrow 3{c^2} - 6c + 2 = 0 \\
  c = \dfrac{{6 \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( {3 \times 2} \right)} }}{{2\left( 3 \right)}} \\
  c = \dfrac{{6 \pm \sqrt {36 - 24} }}{6} \\
  c = \dfrac{{6 \pm \sqrt {12} }}{6} \\
  c = \dfrac{{6 \pm 2\sqrt 3 }}{6} \\
$
Separating the positive terms, we get
$
  c = \dfrac{{6 + 2\sqrt 3 }}{6} \\
  c = 1 + \dfrac{{\sqrt 3 }}{3} \\
  c = 1 + \dfrac{1}{{\sqrt 3 }} \\
$
Separating the negative terms, we get
$
  c = \dfrac{{6 - 2\sqrt 3 }}{6} \\
  c = 1 - \dfrac{{\sqrt 3 }}{3} \\
  c = 1 - \dfrac{1}{{\sqrt 3 }} \\
$
 Thus, $c = 1 \pm \dfrac{1}{{\sqrt 3 }} \in \left[ {0,2} \right]$ and Rolle`s theorem is verified.
Therefore, the value of $c$ is $1 \pm \dfrac{1}{{\sqrt 3 }}$.
So, option B. $1 \pm \dfrac{1}{{\sqrt 3 }}$
Note: All the three conditions must be satisfied to obtain the value of $c$ in Rolle`s theorem. Otherwise the Rolle`s theorem cannot be verified.