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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 $(a, b)$. $3$. $f(a) = f(b)$. Checking option $A$: $f(x) = |\operatorname{sgn}(x)|$ is discontinuous at $x = 0$,so it fails. Checking option $B$: $f(x) = 3x^2 - 2$ on $[2, 3]$. Here $f(2) = 3(4) - 2 = 10$ and $f(3) = 3(9) - 2 = 25$. Since $f(2) 
eq f(3)$,it fails. Checking option $C$: $f(x) = |x - 1|$ on $[0, 2]$. $f(0) = |0 - 1| = 1$ and $f(2) = |2 - 1| = 1$. Thus $f(0) = f(2)$. However,$f(x)$ is not differentiable at $x = 1$,which is inside $(0, 2)$,so it fails. Checking option $D$: $f(x) = x + \frac{1}{x}$ on $[\frac{1}{3}, 3]$. $f(\frac{1}{3}) = \frac{1}{3} + 3 = \frac{10}{3}$ and $f(3) = 3 + \frac{1}{3} = \frac{10}{3}$. Thus $f(\frac{1}{3}) = f(3)$. Since $f(x)$ is a rational function with a non-zero denominator in the interval,it is continuous and differentiable on the interval. Therefore,it satisfies Rolle's theorem.

Which of the following functions satisfies the conditions of Rolle's theorem on the given interval?

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