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(D) Rolle's theorem states that for a function $f(x)$ to satisfy the theorem on $[a, b]$,it must satisfy three conditions: $1$. $f(x)$ is continuous o

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$f(x) = x + \frac{1}{x}$ in $[\frac{1}{3}, 3]$

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(D) Rolle's theorem states that for a function $f(x)$ to satisfy the theorem on $[a, b]$,it must satisfy three conditions: $1$. $f(x)$ is continuous on $[a, b]$. $2$. $f(x)$ is differentiable on $(a, b)$. $3$. $f(a) = f(b)$. Let's check the options: Option $A$: $f(x) = |	ext{sgn}(x)|$ is not continuous at $x = 0$,so it fails. Option $B$: $f(2) = 3(2)^2 - 2 = 10$ and $f(3) = 3(3)^2 - 2 = 25$. Since $f(2) 
eq f(3)$,it fails. Option $C$: $f(x) = |x - 1|$ is not differentiable at $x = 1$,which is in $(0, 2)$,so it fails. Option $D$: $f(x) = x + \frac{1}{x}$ on $[\frac{1}{3}, 3]$. $f(\frac{1}{3}) = \frac{1}{3} + 3 = \frac{10}{3}$. $f(3) = 3 + \frac{1}{3} = \frac{10}{3}$. Since $f(\frac{1}{3}) = f(3)$,and the function is continuous and differentiable in the interval,it satisfies Rolle's theorem.

Which of the following functions can satisfy Rolle's theorem on the given interval?

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