The function f(x) = x^3 - 6x^2 + ax + b satis | vedclass

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(A) For Rolle's theorem to be satisfied on $[1, 3]$,we must have $f(1) = f(3)$.First,calculate $f(1) = (1)^3 - 6(1)^2 + a(1) + b = 1 - 6 + a + b = a +

Vedclass · Accepted Answer

$a = 11, b = -6$

Vedclass · Answer

(A) For Rolle's theorem to be satisfied on $[1, 3]$,we must have $f(1) = f(3)$.First,calculate $f(1) = (1)^3 - 6(1)^2 + a(1) + b = 1 - 6 + a + b = a + b - 5$.Next,calculate $f(3) = (3)^3 - 6(3)^2 + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27$.Equating $f(1) = f(3)$,we get $a + b - 5 = 3a + b - 27$.Subtracting $b$ from both sides: $a - 5 = 3a - 27$.Rearranging terms: $27 - 5 = 3a - a$,which gives $22 = 2a$,so $a = 11$.Rolle's theorem also requires $f'(c) = 0$ for some $c \in (1, 3)$.$f'(x) = 3x^2 - 12x + a = 3x^2 - 12x + 11$.Setting $f'(c) = 0$: $3c^2 - 12c + 11 = 0$.The roots are $c = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = 2 \pm \frac{\sqrt{3}}{3}$.Since $2 - \frac{\sqrt{3}}{3} \approx 2 - 0.577 = 1.423$,which is in $(1, 3)$,the condition is satisfied for $a = 11$.Since $b$ is not uniquely determined by Rolle's theorem conditions alone,we check the options. Option $A$ provides $a = 11$ and $b = -6$.

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. The values of $a$ and $b$ are:

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