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Question: What are the values of \[c\] for which Rolle`s Theorem for the function \(f\left( x \right) = {x^3} ...

What are the values of cc for which Rolle`s Theorem for the function f(x)=x3−3x2+2xf\left( x \right) = {x^3} - 3{x^2} + 2x in the interval [0,2]\left[ {0,2} \right]is verified?
A. c=±1c = \pm 1
B. c=1±13c = 1 \pm \dfrac{1}{{\sqrt 3 }}
C. c=±2c = \pm 2
D. None of these

Explanation

Solution

Hint: Rolles 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 Rolles theorem
Conditions for Rolles 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 Rolles theorem are satisfied.
So, there must be exist c∈[0,2]c \in \left[ {0,2} \right] such that f′(c)=0f'\left( c \right) = 0
f′(c)=3(c)2−6c+2=0f'\left( c \right) = 3{\left( c \right)^2} - 6c + 2 = 0
solving the equation 3(c)2−6c+2=03{\left( c \right)^2} - 6c + 2 = 0 we get
⇒3c2−6c+2=0 c=6±(−6)2−4(3×2)2(3) c=6±36−246 c=6±126 c=6±236  \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=6+236 c=1+33 c=1+13  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=6−236 c=1−33 c=1−13  c = \dfrac{{6 - 2\sqrt 3 }}{6} \\\ c = 1 - \dfrac{{\sqrt 3 }}{3} \\\ c = 1 - \dfrac{1}{{\sqrt 3 }} \\\
Thus, c=1±13∈[0,2]c = 1 \pm \dfrac{1}{{\sqrt 3 }} \in \left[ {0,2} \right] and Rolles 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 Rolles theorem. Otherwise the Rolle`s theorem cannot be verified.