f(x) = |log_e |x|| is differentiable at | vedclass

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(D) The function is defined as $f(x) = |\log_e |x||$.First,consider the domain of the function. The function is undefined at $x = 0$ because $\log_e 0

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$R - \{0, \pm 1\}$

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(D) The function is defined as $f(x) = |\log_e |x||$.First,consider the domain of the function. The function is undefined at $x = 0$ because $\log_e 0$ is undefined.Next,examine the points where the function might not be differentiable. The function $f(x)$ involves an absolute value of a logarithm,which creates sharp corners (cusps) where the argument of the absolute value is zero.Setting $\log_e |x| = 0$,we get $|x| = 1$,which implies $x = 1$ or $x = -1$.At $x = 1$ and $x = -1$,the graph has sharp turns,meaning the derivative does not exist at these points.At $x = 0$,the function has a vertical asymptote,so it is not continuous,and therefore not differentiable.Thus,the function $f(x)$ is differentiable for all real numbers except $x = 0, 1, -1$.This is represented as $R - \{0, 1, -1\}$ or $R - \{0, \pm 1\}$.Therefore,option $(d)$ is correct.

$f(x) = |\log_e |x||$ is differentiable at

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