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Let f(x)=|x-1|+|x+1|.Then f is different...

Let f(x)=|x-1|+|x+1|.Then f is differentiable in:

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|x-1|=x-1,x>1
|x-1|=-(x-1),x<1
|x+1|=x+1,x.-1
|x+1|=-(x+1),x<-1
f(x)=|x-1|+|x-1|
1`lt`x`lt`-1
f(x)=x-a|x+1=2x
f(x)=-(x+1)+x+1=2
...
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