Differentiability of |x| at x = 0

Calculus 1 · Axiom Academy

EXAMPLE Differentiability of |x| at x = 0 Investigating why the absolute value function has a "corner" at the origin Is f(x) = |x| differentiable at x = 0 ? Use the limit definition of the derivative to decide — and reconcile your answer with the fact that |x| is continuous everywhere. The two straight arms meet at a sharp point. To the left the slope is constant at -1 ; to the right it is +1 . Because the two slopes disagree at x = 0 , no single tangent line fits there. Nice work — you've shown that |x| is not differentiable at x = 0 . Here's what we learned: One-sided limits must agree: for a derivative to exist at a point, both one-sided limits of the difference quotient must exist and be equal. Here they are +1 and -1 . The corner is the tell: the sharp point at x = 0 means the tangent line is not well-defined — approaching from the left gives slope -1 , from the right gives slope +1 . Continuous but not differentiable: |x| is continuous everywhere ( ), yet it fails to be differentiable at 0 . Continuity does not guarantee differentiability. This is the classic counterexample: a continuous function need not be differentiable at every point.

This is the written version of the interactive lesson above. See the full Calculus 1 course.