Continuous does not mean differentiable: you can draw it without lifting the pen and still have no tangent line

01The mistake

Students assert that a continuous function is differentiable everywhere. Shown $f(x)=|x|$, which is continuous at 0 and has no derivative there, they either deny it is continuous or deny that the derivative fails — both to preserve the equivalence.

The tell is asking for the direction of the implication in words. “If a function is differentiable at a point, what else do you know? If it is continuous at a point, what else do you know?” The first has an answer and the second does not. Students holding the biconditional give the same answer to both, which is the whole diagnosis in one question.

It makes the three failure modes invisible. Corners, cusps and vertical tangents are all continuous and non-differentiable, and a student who believes continuity is sufficient has no reason to look for any of them. That is U2-CA6, and it is a direct consequence rather than a separate gap.

The cost lands hardest in Unit 5. The Mean Value Theorem requires differentiability on the open interval, and Rolle's requires it too; a student who reads continuity as enough will apply MVT to $|x|$ on $[-1,1]$ and conclude something false. Hypothesis-checking questions are common on the exam precisely because this belief is common.

02Why it makes sense to the student

Every function in the course before this point was differentiable wherever it was continuous. Polynomials, exponentials, sines, rational functions on their domains — the entire library. The equivalence holds across all of a student's experience, so it is an excellent empirical generalisation from the available data.

The one-directional implication is genuinely easy to garble. “Differentiability implies continuity” is a sentence about direction, and directions are the first thing lost when a fact is stored as an association between two words. Students remember that the words go together and lose which one comes first.

The pen-lifting metaphor is doing quiet damage. “Continuous means you can draw it without lifting your pen” is a useful image, and it makes continuity feel like smoothness. A corner is drawn without lifting the pen and is not smooth, but nothing in the metaphor distinguishes the two.

And derivatives are introduced procedurally. Students spend Unit 2 differentiating things that always work, so the question of whether a derivative exists never arises as a live question. The power rule does not ask permission.

03The correction

Put the implication on the board with an arrow, in one direction only: differentiable $\Rightarrow$ continuous. Then write the converse with a line through it. Making the asymmetry visual is worth more than saying it, because the failure mode is specifically losing the direction.

Then give the counterexample immediately and use it all year. $f(x) = |x|$ is continuous at 0 — the limit and the value both equal 0 — and the derivative does not exist there, because the left-hand difference quotient goes to $-1$ and the right-hand one goes to $+1$. Compute both one-sided limits explicitly; the disagreement is the non-existence, and students who have seen it computed stop treating it as an assertion.

Name the three ways differentiability fails at a point where the function is continuous, so students have something to look for: a corner (one-sided derivatives disagree, like $|x|$), a cusp (they diverge to opposite infinities, like $x^{2/3}$), and a vertical tangent (the derivative diverges, like $x^{1/3}$). Four failure modes total if you include an actual discontinuity.

Kill the smoothness reading of the pen metaphor directly. Continuity means no gaps. Differentiability means no gaps and no sharp turns. The pen picture captures the first and says nothing about the second, and students should be told the metaphor is incomplete rather than left to discover it.

A useful classroom test: “Give an example of a function that is continuous at a point but not differentiable there. Now give one that is differentiable but not continuous.” The first has many answers; the second has none, and knowing why it has none is the concept. A student who produces an example for the second question has the implication backwards.

04A sample question

05What each wrong answer reveals

• A Continuity taken as sufficient. The justification names the pen-lifting picture directly — the graph is unbroken — and treats it as settling both questions. This is the misconception and the metaphor that produced it, visible in the same sentence. Compute the two one-sided difference quotients in front of them; the $-1$ and $+1$ do what no explanation does.
• B Correct. $|x|$ is continuous at 0 since $\lim_{x\to 0}|x| = 0 = |0|$. It is not differentiable there because the left-hand derivative is $-1$ and the right-hand derivative is $+1$, so the limit defining $f'(0)$ does not exist.
• C The implication inverted. This student has the relationship backwards in the strongest possible way, asserting something that cannot happen for any function. Worth addressing directly rather than gently: differentiability requires continuity, so “differentiable but not continuous” is not a description of an unusual function, it is impossible. That framing tends to stick.
• D Non-differentiability read as discontinuity. The student correctly noticed the sharp turn — the observation that matters — and concluded the function must also fail continuity, because in their model the two travel together. Closer than A: they have seen the feature and mis-assigned its consequence. Ask them to evaluate $\lim_{x\to 0}|x|$ and $|0|$ and compare.

A and D are the same equivalence failing in opposite directions, and D is the more promising: that student is looking at the corner, which A is not. C is a separate and more serious error, since it asserts an impossibility. The two-part question above — give me an example each way — sorts all three faster than any item, because the impossible half is where the direction of the implication lives.

06Try it in Mistake Master