Connecting differentiability and continuity — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The theorem that differentiability implies continuity, the false converse that continuity does not imply differentiability, identification of four types of non-differentiable points, and testing for differentiability at junctions of piecewise-defined functions.

You should already know: Definition of continuity at a point via limits, Definition of the derivative as a limit of difference quotients, Evaluating one-sided limits for piecewise functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Connecting differentiability and continuity?

This topic explores the formal logical relationship between two core properties of functions at a point: continuity (the function has no break, hole, or jump at the point) and differentiability (the derivative exists at the point, meaning a unique, finite tangent line can be drawn there). Per the AP Calculus CED, this subtopic falls within Unit 2: Differentiation: Definition and Fundamental Properties, which accounts for 10–12% of the total AP exam score, with this topic making up roughly 3–4% of the total exam. It appears in both multiple-choice (MCQ) questions as a standalone conceptual check, and in free-response (FRQ) questions as a required justification step for piecewise function problems. This topic is not just a theoretical result: it establishes the ground rule that any point where a function is discontinuous cannot be differentiable, which simplifies checking for derivative existence in almost all future problems. Synonyms for differentiability at a point include “the derivative exists at $x=a$”; notation follows standard conventions: $f^{\prime }(a)$ exists if and only if $f$ is differentiable at $x=a$.

2. Theorem: Differentiability Implies Continuity

The core result of this topic is a formal logical implication: If a function $f$ is differentiable at $x=a$, then $f$ must be continuous at $x=a$. We can prove this quickly using limit laws. If $f$ is differentiable at $a$, the derivative exists as a finite limit: $$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$ To confirm continuity, we need to show ${\lim }_{h\to 0}\left[f(a+h)-f(a)\right]=0$, which is the requirement for continuity at $a$. Rewrite the limit using the derivative expression: $$\underset{h\to 0}{\lim }\left[f(a+h)-f(a)\right]=\underset{h\to 0}{\lim }\left(\frac{f(a+h)-f(a)}{h}\cdot h\right)$$ By the product rule for limits, this splits into ${\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot {\lim }_{h\to 0}h=f^{\prime }(a)\cdot 0=0$, which confirms continuity. The contrapositive of this theorem is even more useful for problem-solving: If $f$ is not continuous at $x=a$, then $f$ is not differentiable at $x=a$. This lets us stop testing for differentiability immediately if we find a discontinuity, saving time on the exam.

Worked Example

Problem: Let $f(x)=\{\begin{array}{ll}3x-2& x<2\\ 3x+1& x\ge 2\end{array}$. Is $f$ differentiable at $x=2$? Justify your answer.

1. First check continuity at $x=2$. Calculate the left-hand limit: ${\lim }_{x\to 2^{-}}f(x)=3(2)-2=4$.
2. Calculate the right-hand limit and function value: ${\lim }_{x\to 2^{+}}f(x)=f(2)=3(2)+1=7$.
3. Since the two-sided limit does not equal $f(2)$ (the left limit is $4\ne 7$), $f$ is discontinuous at $x=2$.
4. By the contrapositive of the differentiability-implies-continuity theorem, $f$ cannot be differentiable at $x=2$.

Exam tip: Always check for discontinuity first when testing differentiability. If you find a discontinuity, you can stop immediately and write your conclusion, saving 1-2 minutes on exam day.

3. Continuity Does Not Imply Differentiability

While differentiability guarantees continuity, the converse is not true: continuity is a necessary, but not sufficient, condition for differentiability. A function can be fully continuous at $x=a$ but still fail to be differentiable there, most commonly in four cases: (1) corners (left-hand derivative ≠ right-hand derivative), (2) cusps (one-sided derivatives approach opposite infinities), (3) vertical tangents (the derivative approaches ±infinity, so it is not finite), and (4) oscillating tangents (the limit of the difference quotient does not exist). On the AP exam, the most common test of this idea is for piecewise functions that are continuous at the junction point, so you need to check if the left-hand derivative equals the right-hand derivative. For smooth pieces of a piecewise function, the one-sided derivative at the junction is just the derivative of the piece evaluated at the junction point. If the two one-sided derivatives are equal and finite, the derivative exists; otherwise, it does not.

Worked Example

Problem: Let $f(x)=\{\begin{array}{ll}{x}^2+3& x\le 3\\ 7x-9& x>3\end{array}$. Is $f$ differentiable at $x=3$? Justify your answer.

1. First confirm continuity at $x=3$: $f(3)=3^2+3=12$. The right-hand limit is $7(3)-9=12$, so $f$ is continuous at $x=3$.
2. Calculate the left-hand derivative: the derivative of $x^2+3$ is $2x$, so $f_{-}^{\prime }(3)=2(3)=6$.
3. Calculate the right-hand derivative: the derivative of $7x-9$ is $7$, so $f_{+}^{\prime }(3)=7$.
4. Since $f_{-}^{\prime }(3)=6\ne 7=f_{+}^{\prime }(3)$, the two-sided derivative does not exist, so $f$ is continuous but not differentiable at $x=3$.

Exam tip: Do not just compare derivatives of the pieces and skip the continuity check. If the function is discontinuous, equal derivatives on each side do not make it differentiable.

4. Identifying Non-Differentiable Points From Graphs

A common AP exam question gives you the graph of $f(x)$ and asks you to count how many non-differentiable points exist in a given interval, or identify which statement about differentiability is true. To solve these, you just check for the four types of non-differentiable points directly from the graph: any discontinuity, corner, cusp, or vertical tangent is non-differentiable. The intuitive rule to remember is: if you cannot draw a single unique tangent line with finite slope at the point, the function is not differentiable there. Holes, jumps, and infinite discontinuities are all obvious discontinuities, but don't miss corners where two lines meet at different slopes, or points where the graph flattens to a vertical line.

Worked Example

Problem: The graph of $y=f(x)$ on the open interval $(-4,4)$ has a jump discontinuity at $x=-2$, a corner at $x=0$, a vertical tangent at $x=2$, and is smooth at all other points. How many non-differentiable points are in $(-4,4)$?

1. List all candidate points: $x=-2$, $x=0$, $x=2$, all inside the open interval.
2. $x=-2$ is a discontinuity: by the contrapositive theorem, it is non-differentiable (count = 1).
3. $x=0$ is a corner: it is continuous, but the left slope does not equal the right slope, so it is non-differentiable (count = 2).
4. $x=2$ has a vertical tangent: the slope is infinite, so it is not a finite derivative, hence non-differentiable (count = 3).
5. There are no other candidate points, so the total number of non-differentiable points is 3.

Exam tip: Don't forget to count vertical tangents as non-differentiable. Most students remember corners and discontinuities but miss vertical tangents on graph problems.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After confirming the derivatives of the two pieces of a piecewise function are equal at $x=a$, conclude the function is differentiable without checking continuity first. Why: Students assume matching derivatives imply the function is connected, so they skip the required continuity check. Correct move: Always check continuity at $x=a$ before comparing one-sided derivatives. If discontinuous, stop and conclude non-differentiability.
• Wrong move: Conclude a function is not continuous at $x=a$ just because it is not differentiable at $x=a$. Why: Students confuse the direction of the implication, incorrectly inverting the "differentiability implies continuity" rule. Correct move: Remember only "discontinuous → not differentiable" is always true; non-differentiability does not imply discontinuity, so always test continuity separately.
• Wrong move: Count a removable discontinuity (hole) as differentiable because the limit of the function exists at the point. Why: Students confuse the limit of the function with the function being continuous at the point. Correct move: Any discontinuity (removable, jump, infinite) means the function is not continuous, hence not differentiable, regardless of whether the limit exists.
• Wrong move: Conclude $f(x)=|x|$ is not continuous at $x=0$ because it is not differentiable at $x=0$. Why: Students forget that $|x|$ is the classic example of a continuous non-differentiable function, and mix up the implication direction. Correct move: Check continuity first, then check differentiability: $|x|$ is continuous at 0, but non-differentiable.
• Wrong move: When comparing one-sided derivatives for a piecewise junction at $x=a$, evaluate the derivatives at $x=0$ instead of $x=a$. Why: Students rush and default to plugging in the most common junction point, leading to an incorrect comparison. Correct move: Explicitly label the junction $x=a$, write the derivative of each piece, and evaluate at $x=a$ before comparing.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Let $f$ be a function defined for all real numbers such that ${\lim }_{x\to 3}f(x)=5$ and $f(3)=7$. Which of the following statements must be true? A) $f$ is continuous at $x=3$, and $f$ is differentiable at $x=3$ B) $f$ is continuous at $x=3$, but $f$ is not differentiable at $x=3$ C) $f$ is not continuous at $x=3$, but $f$ is differentiable at $x=3$ D) $f$ is not continuous at $x=3$, and $f$ is not differentiable at $x=3$

Worked Solution: By definition, a function is continuous at $x=a$ if and only if ${\lim }_{x\to a}f(x)=f(a)$. For $a=3$, we have ${\lim }_{x\to 3}f(x)=5\ne 7=f(3)$, so $f$ is not continuous at $x=3$. By the contrapositive of the theorem that differentiability implies continuity, a discontinuous function cannot be differentiable at that point. Therefore, $f$ is also not differentiable at $x=3$. The correct answer is D.

Question 2 (Free Response)

Let $g(x)=\{\begin{array}{ll}a{x}^3+bx& x\le 1\\ 4x^2-2& x>1\end{array}$. (a) Find all values of $a$ and $b$ such that $g(x)$ is continuous at $x=1$. (b) For values of $a$ and $b$ that make $g$ continuous at $x=1$, when is $g$ also differentiable at $x=1$? Justify your answer. (c) Suppose $a=3$ and $b=-1$. Is $g$ differentiable at $x=1$? Justify.

Worked Solution: (a) For continuity at $x=1$, ${\lim }_{x\to 1^{-}}g(x)={\lim }_{x\to 1^{+}}g(x)=g(1)$. The right-hand limit is ${\lim }_{x\to 1^{+}}(4x^2-2)=4(1)-2=2$. The left-hand limit is ${\lim }_{x\to 1^{-}}(a{x}^3+bx)=a+b=g(1)$. All $a,b$ satisfying $\boxed{a+b=2}$ make $g$ continuous at $x=1$. (b) To test differentiability for continuous $g$, compare left and right derivatives. The right-hand derivative is $g_{+}^{\prime }(1)=\frac{d}{dx}(4x^2-2){|}_{x=1}=8x{|}_{x=1}=8$. The left-hand derivative is $g_{-}^{\prime }(1)=\frac{d}{dx}(a{x}^3+bx){|}_{x=1}=3a{x}^2+b{|}_{x=1}=3a+b$. Set equal for differentiability: $3a+b=8$. Substitute $b=2-a$ from (a): $3a+2-a=8⟹2a=6⟹a=3,b=-1$. $g$ is differentiable at $x=1$ only when $a=3,b=-1$; for all other continuous $g$, it is continuous but not differentiable at $x=1$. (c) For $a=3,b=-1$, we confirm $a+b=3-1=2$, so $g$ is continuous at $x=1$. We already found $g_{-}^{\prime }(1)=3(3)+(-1)=8$, and $g_{+}^{\prime }(1)=8$, so the derivatives are equal. Therefore, $g$ is differentiable at $x=1$.

Question 3 (Application / Real-World Style)

A city models the depth of water in a reservoir, in meters, $d$ days after the start of a month, as $D(d)=\{\begin{array}{ll}12+0.2d^2& 0\le d\le 10\\ 5d-13& d>10\end{array}$. Water resource managers require the depth function to be differentiable at all $d>0$ to calculate a well-defined rate of change of depth for flood planning. Is the model differentiable at $d=10$? Justify your answer, and explain what the result means for flood planning.

Worked Solution: First check continuity at $d=10$: Left-hand depth is $D(10)=12+0.2(10{)}^2=12+20=32$ meters. Right-hand limit is $5(10)-13=50-13=37$ meters. Since $32\ne 37$, $D(d)$ is discontinuous at $d=10$. By the theorem differentiability implies continuity, $D(d)$ is not differentiable at $d=10$. In context, this means the model predicts a sudden 5-meter jump in reservoir depth at 10 days, so there is no well-defined rate of depth change at that point, which means the model cannot be used for reliable flood planning at the 10-day mark.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all subsequent differentiation work in AP Calculus BC. Next, you will learn the product rule, quotient rule, and chain rule for differentiating combinations of functions, all of which assume that the underlying functions are differentiable on their domains. You will also apply this relationship when you study derivatives of inverse functions, implicit differentiation, and the Mean Value Theorem, which explicitly requires a function to be continuous on a closed interval and differentiable on the open interval to apply. Without mastering the order of checking (continuity first, then differentiability) and the one-way nature of the implication, you will lose easy points on justification questions, which are graded strictly for correct logical reasoning on the AP exam. Follow-on topics you will connect to this one include: Product and quotient differentiation rules Chain rule for composite functions Mean Value Theorem for derivatives