Ask Questions, Get Answers


The function given by $y=||x|-1|$ is differentiable for all real numbers except the points

$(a)\;\{0,1,-1\}\qquad(b)\;\pm 1\qquad(c)\;1\qquad(d)\;-1$

1 Answer

Given function is $y=||x|-1|$
$y=\left\{\begin{array}{1 1}-|x|+1&if\;|x| < 1\\|x|-1&if\;|x| \geq 1\end{array}\right.$
$\;\;\;=\left\{\begin{array}{1 1}-|x|+1&if\;-1 < x < 1\\|x|-1&if\;x \geq -1\;or\;x\geq 1\end{array}\right.$
$\;\;\;=\left\{\begin{array}{1 1}-x-1&if\;x \leq -1\\x+1&if\;-1 < x < 0 \\ -x+1&if\;0\leq x < 1\\x-1&if\;x \geq 1\end{array}\right.$
Here $Ly'(-1)=-1$ and $Ry'(-1)=1$
$Ly(0)=1$ $Ry'(0)=-1$
and $Ly'(1)=-1$ and $Ry'(1)=1$
$\Rightarrow y$ is not differentiable at $x=-1,0,1$
Hence (a) is the correct option.
answered Jan 3, 2014 by sreemathi.v

Related questions