# Chapter 3 - Differentiation - 3.2 The Derivative as a Function - Exercises - Page 116: 75

The function $f(x)=|x^2-1|$ is not differentailble at $x=\pm1$.

#### Work Step by Step

Since $$|x^2-1|=0\Longrightarrow x^2-1=0\Longrightarrow (x+1)(x-1)=0.$$ The function $f(x)=|x^2-1|$ has corners at $x=\pm 1$, hence it is not differentailble at $x=\pm1$.

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