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i solved for f'(x) and i got $\frac{x-1}{\left|x-1\right|}$. i think that this is defined for 1>x>1. as $\frac{x-1}{\left|x-1\right|}$ can't be defined for x=1 because the denominator will be 0. I don't know if this is correct

PiGamma
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john.r
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    It is defined for $\color{red}{x\ne1}$. Indeed, it is also known as $\operatorname{sgn}(x-1)$, and the sign of $0$ is not defined. – Bernard Nov 26 '17 at 15:01

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You have $$f’(x)=\frac{x-1}{|x-1|}.$$ For $x>1$ you have that $|x-1|=x-1$, hence $$f’(x)=\frac{x-1}{x-1}=1.$$ For $x<1$ you have $|x-1|=-(x-1)$ and therefore $$f’(x)=\frac{x-1}{-(x-1)}=-1.$$ Finally, for $x=1$ the function is undefined.

James Garrett
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