Why does differentiability imply continuity?
You will be able to: Distinguish a necessary continuity condition from sufficient differentiability.
BC foundation: Unit 2 shares its derivative foundations with AB. Connect each rule to rates, graphs and its domain conditions. Use the written challenges to explain your reasoning. Chain-rule compositions, implicit and inverse differentiation, and advanced applications come later.
Why does differentiability imply continuity?
A smoothly drawn path cannot suddenly jump at the point where it has a finite local slope. But a connected path can still turn sharply, like the bottom of a V.
A useful starting point: How do you estimate a derivative from a graph? →
Words and symbols before equations
- Differentiable
- Having a finite derivative at the specified point.
- Continuous
- The function value agrees with its nearby limit.
- Necessary condition
- A condition that must hold, but may not be enough.
- Converse
- A statement formed by reversing an implication.
What this picture assumes
Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. Ordinary finite derivatives at interior inputs. Continuity is necessary, not sufficient.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- At zero, f=x² is continuous and has derivative 0.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
For h≠0, f(a+h)−f(a)=([f(a+h)−f(a)]/h)h. If the quotient tends to finite f′(a), this product tends to zero, so nearby values approach f(a). That is continuity.
Thus differentiable implies continuous. Its contrapositive is useful: a discontinuity guarantees no derivative at that point.
The converse fails. For f(x)=abs(x), f is continuous at 0, but the right quotient is 1 and the left quotient is −1; no common derivative exists.
Checking continuity is the first step, not the entire test. If continuity passes, examine the difference quotient’s one-sided finite limits.
| Question | Continuous at a | Differentiable at a |
|---|---|---|
| Required agreement | Nearby heights and f(a) agree | Finite side difference-quotient limits agree |
| Corner abs(x) at 0 | Yes | No |
| Logical relationship | Necessary for differentiability | Sufficient for continuity |
A worked example, step by step
Classify f(x)=x² for x≠1 with f(1)=5 at x=1.
- The nearby limit of x² at 1 is 1.
- The assigned value is 5.
- These differ, so f is not continuous at 1.
- A differentiable function would be continuous; therefore f′(1) does not exist.
Continuous does not mean differentiable. A visually connected curve may have a corner or other derivative failure.
If a derivative does not exist, must the function be discontinuous?
Compare with an explanation
No. The V-shaped absolute-value function is a continuous counterexample.
Predict. Change one thing. Explain.
Switch between a smooth parabola, a V and a changed point. Check continuity first, then compare the local slopes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At zero, f=x² is continuous and has derivative 0.
Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. Ordinary finite derivatives at interior inputs. Continuity is necessary, not sufficient.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using a difference quotient, tangent slope, or derivative rule with its domain conditions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionUse f(x)=abs(x) to refute “every continuous function is differentiable.” Include both one-sided quotients.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The limit and f(0) both equal 0, so it is continuous.
- 1 point: For h>0, abs(h)/h=1.
- 1 point: For h<0, abs(h)/h=−1.
- 1 point: The finite side limits differ, so f′(0) does not exist.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does differentiability guarantee?
Continuity at that point.
RECALL 2What does continuity fail to guarantee?
Existence of a finite derivative.
RECALL 3What is a useful counterexample?
abs(x) at x=0.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does differentiability imply continuity?
- Differentiable at a ⇒ continuous at a.
- Not continuous at a ⇒ not differentiable at a.
- Continuous alone is inconclusive.
Remember: Continuous does not mean differentiable. A visually connected curve may have a corner or other derivative failure.
Conditions: Original equation-driven model; readouts are rounded. Graphs have labeled linear scales and finite sampled windows; algebra supplies exact conclusions. Ordinary finite derivatives at interior inputs. Continuity is necessary, not sufficient.
Refresh Kid · AP Calculus BC Unit 2 · Objectives FUN-2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.4, FUN-2.A. CED effective Fall 2020, with Fall 2026 clarifications, checked September 17, 2026. Unit 2 has 10 official topics. Focused lesson titles and questions are original Refresh Kid teaching material.
Derivative definitions require finite two-sided limits at interior inputs. Graphs and finite tables supply evidence and estimates, not universal proofs. Continuity is necessary but insufficient for differentiability. Rules retain their original domain restrictions, and trigonometric inputs use radians. Chain-rule compositions, implicit differentiation, inverse-function differentiation and L’Hôpital’s rule are deferred to later units. Piecewise joins are checked for continuity before their side slopes are compared.
The Organic Chemistry Tutor video titles, creators and descriptions were checked; full videos were not reviewed. Khan Academy’s unit destination was checked; its JavaScript lesson content was not fully readable by the research tool. OpenStax Sections 3.1, 3.3 and 3.5 were consulted for conceptual cross-checking. No creator scripts, questions, diagrams or artwork were copied. Refresh Kid is not endorsed by these providers.
GitHub’s 3D website collection and its camera-control example informed optional spatial inspection. Graph geometry is generated from the stated original equations; self-hosted Three.js retains its MIT license. The optional 3D model uses separate planes for f and f′. Depth distinguishes panels, not an additional variable; camera rotation does not change their values. Use the complete labeled 2D graphs and text to read precise coordinates and slopes.
Independent teacher review and observation of students remain pending. Checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned to this lesson.
Learn → Explore → Practice → Review is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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