For which interval does the function f(x) = f | vedclass

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(D) Rolle's theorem requires three conditions for a function $f(x)$ on an interval $[a, b]$:$1$. $f(x)$ must be continuous on $[a, b]$.$2$. $f(x)$ mus

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For no interval

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(D) Rolle's theorem requires three conditions for a function $f(x)$ on an interval $[a, b]$:$1$. $f(x)$ must be continuous on $[a, b]$.$2$. $f(x)$ must be differentiable on $(a, b)$.$3$. $f(a) = f(b)$.Given $f(x) = \frac{x^2 - 3x}{x - 1}$.The function is undefined at $x = 1$. Therefore,it is discontinuous at $x = 1$.For any interval $[a, b]$ containing $x = 1$,the function fails the continuity condition.In the given options:- For $[0, 3]$,$1 \in (0, 3)$,so it is not continuous.- For $[-3, 0]$,the function is continuous and differentiable,but $f(-3) = \frac{9+9}{-4} = -4.5$ and $f(0) = 0$. Since $f(-3) 
eq f(0)$,Rolle's theorem does not apply.- For $[1.5, 3]$,the function is continuous and differentiable,but $f(1.5) = \frac{2.25 - 4.5}{0.5} = -4.5$ and $f(3) = \frac{9 - 9}{2} = 0$. Since $f(1.5) 
eq f(3)$,Rolle's theorem does not apply.Thus,there is no interval among the choices where all conditions are satisfied.

For which interval does the function $f(x) = \frac{x^2 - 3x}{x - 1}$ satisfy all the conditions of Rolle's theorem?

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