Taylor Polynomials: Why Abs(0) Doesn't Have Center at Xo=0

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Why does f(x) = abs (0) not have the first Taylor polynomial center at Xo = 0 ?
Does it have second Taylor polynomial center at Xo = 0 ?
 
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ado1lz6 said:
Why does f(x) = abs (0) not have the first Taylor polynomial center at Xo = 0 ?
Does it have second Taylor polynomial center at Xo = 0 ?

What is the derivative of |x| at x= 0?
 
That only explains that f(x) does not have second Taylor polynomial center at Xo=0. What about the first Taylor polynomial ?
Thanks a lot for reply
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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