Mistake Master · AP Calculus · Unit 2 · Step-Through Animation
Smooth Is a Stronger Word Than Unbroken
You'll learnwhy a derivative forces continuity, why the sentence is false read backwards, and how to name which of the four failures a graph is actually showing.
Topic 1.11 asked whether a graph is unbroken at a point. This one asks a strictly harder question: whether it is also smooth there. If f′(a) exists then f is continuous at a — two lines of algebra settle it, and the useful form is the contrapositive: a break at a rules out f′(a) with no further work. Then comes the sentence read backwards, and it is false. The counterexample is small: the V of f(x) = |x| has no break at all, and its difference quotient still returns +1 from one side and −1 from the other. Two numbers, one limit, no derivative. Four features do this, and telling a cusp from a vertical tangent is where the marks go.
differentiable ⇒ continuous · the reverse arrow is false
Before you start
What you're looking at
Two boxes and two arrows: differentiable at a, continuous at a, and the claim that each one implies the other. Only one of those arrows survives.
The question
If a graph has no break at a point, is that enough for a derivative to exist there — and if not, what exactly goes wrong?
Watch for
The reverse arrow being struck out, the two rise-over-run triangles that never agree, and the gallery where a cusp and a vertical tangent stop looking alike.
Mistake Master