If the function f(x) = ax^3 - bx^2 - 8x - 4 s | vedclass

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(A) Step $1$: Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$, we must have $f(1) = f(3)$.Step $2$: Calculate $f(1) = a(1)^3 - b(1)^2 - 8(1) - 4 =

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$2$

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(A) Step $1$: Since $f(x)$ satisfies Rolle's theorem on $[1, 3]$, we must have $f(1) = f(3)$.Step $2$: Calculate $f(1) = a(1)^3 - b(1)^2 - 8(1) - 4 = a - b - 12$.Step $3$: Calculate $f(3) = a(3)^3 - b(3)^2 - 8(3) - 4 = 27a - 9b - 24 - 4 = 27a - 9b - 28$.Step $4$: Equating $f(1) = f(3)$, we get $a - b - 12 = 27a - 9b - 28$, which simplifies to $26a - 8b = 16$, or $13a - 4b = 8$ (Equation $1$).Step $5$: Find $f'(x) = 3ax^2 - 2bx - 8$. Given $f'(2) = 0$, we have $3a(2)^2 - 2b(2) - 8 = 0$, which is $12a - 4b = 8$ (Equation $2$).Step $6$: Subtracting Equation $2$ from Equation $1$: $(13a - 4b) - (12a - 4b) = 8 - 8$, which gives $a = 0$.Step $7$: Substituting $a = 0$ into Equation $2$: $12(0) - 4b = 8$, so $-4b = 8$, which gives $b = -2$.Step $8$: Therefore, $a - b = 0 - (-2) = 2$.

If the function $f(x) = ax^3 - bx^2 - 8x - 4$ satisfies Rolle's theorem in the interval $[1, 3]$, and $f'(2) = 0$, then the value of $a - b$ is:

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