The points at which the function f(x) = |x - | vedclass

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Question

(D) function $f(x)$ is not differentiable at points where the expression inside the absolute value is zero,or where the function itself is discontinuo

Vedclass · Accepted Answer

$0.5, 1, \frac{\pi}{2}$

Vedclass · Answer

(D) function $f(x)$ is not differentiable at points where the expression inside the absolute value is zero,or where the function itself is discontinuous.$1$. The absolute value functions $|x - 0.5|$ and $|x - 1|$ are not differentiable at $x = 0.5$ and $x = 1$ respectively.$2$. The function $	an x$ is not defined (and thus not differentiable) at $x = \frac{\pi}{2} + n\pi$. Within the interval $(0, 2)$,$\frac{\pi}{2} \approx 1.57$,which lies in the interval.$3$. Combining these,the function $f(x)$ is not differentiable at $x = 0.5$,$x = 1$,and $x = \frac{\pi}{2}$.Thus,the correct option is $D$.

The points at which the function $f(x) = |x - 0.5| + |x - 1| + \tan x$ does not have a derivative in the interval $(0, 2)$ are:

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