Since, for domain of \(\sqrt{1-(x+1)^2},\) we have
1 - (x + 1)2 \(\geq\) 0
⇒ (x + 1)2 \(\leq\) 1
⇒ -1 \(\leq\) x + 1 \(\leq\) 1
⇒ -2 \(\leq\) x \(\leq\) 0
For domain of \(\sqrt{1-(x-1)^2},\) we have
1 - (x - 1)2 \(\geq\) 0
⇒ (x - 1)2 \(\leq\) 1
⇒ -1 \(\leq\) x - 1 \(\leq\) 1
⇒ 0 \(\leq\) x \(\leq\) 2
Case I: If x \(\in\) [-2, 0] then \(\sqrt{1-(x-1)^2},\) is not defined and
\(-\sqrt{1-(x+1)^2}\leq0\)
but |x| \(\geq\) 0
\(\therefore\) Min {\(-\sqrt{1-(x+1)^2},\) \(\sqrt{1-(x-1)^2}\), |x|} = \(-\sqrt{1-(x+1)^2}\) if x \(\in\) [-2, 0]
Case II: If x \(\in\) [0, 1] then \(-\sqrt{1-(x+1)^2}\) is not defined
And |x| \(\leq\) \(\sqrt{1-(x-1)^2}\)
\(\therefore\) Min {\(-\sqrt{1-(x+1)^2},\) \(\sqrt{1-(x-1)^2}\), |x|} = |x| if x \(\in\) [0, 1]
Case III: If x \(\in\) [1, 2] then \(-\sqrt{1-(x+1)^2}\) is not defined
and |x| \(\geq\) \(\sqrt{1-(x-1)^2}\) (\(\because\) Maximum value of \(\sqrt{1-(x-1)^2}\) is 1)
\(\therefore\) Min {\(-\sqrt{1-(x+1)^2}\), \(\sqrt{1-(x-1)^2}\), |x|} = \(\sqrt{1-(x-1)^2}\) , if x \(\in\)[1, 2]
\(\therefore\) f(x) = \(\begin{cases}-\sqrt{1-(x+1)^2}&;x\in[-2,0]\\|x|&;x\in[0,1]\\ \sqrt{1-(x-1)^2}&x\in[1,2]\end{cases}\)
and its graph is
Total number of non-differentials points is 2 which is x = 0, x = 1
