If the function f(x) = x^3 - 6x^2 + ax + b de | vedclass

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(C) For Rolle's theorem to hold on $[1, 3]$,we must have $f(1) = f(3)$.$f(1) = 1^3 - 6(1)^2 + a(1) + b = a + b - 5$.$f(3) = 3^3 - 6(3)^2 + a(3) + b =

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$a = 11, b \in R$

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(C) For Rolle's theorem to hold on $[1, 3]$,we must have $f(1) = f(3)$.$f(1) = 1^3 - 6(1)^2 + a(1) + b = a + b - 5$.$f(3) = 3^3 - 6(3)^2 + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27$.Equating $f(1) = f(3)$ gives $a + b - 5 = 3a + b - 27$,which simplifies to $2a = 22$,so $a = 11$.Since $b$ cancels out in the equation $f(1) = f(3)$,$b$ can be any real number $(b \in R)$.Also,$f'(x) = 3x^2 - 12x + a = 3x^2 - 12x + 11$.By Rolle's theorem,$f'(c) = 0$ for $c = \frac{2\sqrt{3} + 1}{\sqrt{3}} = 2 + \frac{1}{\sqrt{3}}$.$f'(c) = 3(2 + \frac{1}{\sqrt{3}})^2 - 12(2 + \frac{1}{\sqrt{3}}) + 11 = 3(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}) - 24 - \frac{12}{\sqrt{3}} + 11 = 12 + 4\sqrt{3} + 1 - 24 - 4\sqrt{3} + 11 = 0$.Thus,the condition $f'(c) = 0$ is satisfied for $a = 11$ and any $b \in R$.

If the function $f(x) = x^3 - 6x^2 + ax + b$ defined on $[1, 3]$ satisfies Rolle's theorem for $c = \frac{2\sqrt{3} + 1}{\sqrt{3}}$,then:

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