The function y = |sin x| is continuous for an | vedclass

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(D) The function $f(x) = |\sin x|$ is continuous for all real numbers $x$. However,it is not differentiable at points where the graph has sharp corner

Vedclass · Accepted Answer

$x = k\pi$ ($k$ is an integer)

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(D) The function $f(x) = |\sin x|$ is continuous for all real numbers $x$. However,it is not differentiable at points where the graph has sharp corners (cusps). From the graph of $f(x) = |\sin x|$,we can observe that the function has sharp corners at $x = 0, \pm \pi, \pm 2\pi, \dots$. These points are of the form $x = k\pi$,where $k$ is any integer. Therefore,the function is not differentiable at $x = k\pi$ for any integer $k$.

The function $y = |\sin x|$ is continuous for any $x$,but it is not differentiable at:

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