# Connecting differentiability and continuity

> AP Calculus AB · Differentiation: Definition and Fundamental Properties
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u2-connecting-differentiability-and-continuity/

This subtopic explores the logical relationship between differentiability and continuity for functions, including the core theorem that differentiability implies continuity and why the converse is false. It covers identifying non-differentiable points and working with piecewise functions for AP Calculus AB.

**Prerequisites:** [Limit definition of the derivative at a point](https://www.owlsprep.com/study/ap-calculus-ab-limit-definition-derivative/); [Definition of continuity at a point](https://www.owlsprep.com/study/ap-calculus-ab-continuity-definition/); [Evaluating one-sided limits](https://www.owlsprep.com/study/ap-calculus-ab-one-sided-limits/)

## Learning objectives

- Explain the logical relationship between differentiability and continuity
- Prove that differentiability implies continuity
- Identify non-differentiable points from graphs and functions
- Solve for constants to make piecewise functions differentiable

## The Core Theorem: Differentiability Implies Continuity

The first and most important result for this topic is the theorem that formalizes the relationship between the two properties: *If a function $f$ is differentiable at $x=a$, then $f$ must be continuous at $x=a$*. We can prove this directly from the definitions of both concepts.

**Derivation:** Prove that differentiability implies continuity

*Starting from:* If $f$ is differentiable at $x=a$, $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists as a finite limit. We need to show $\lim_{h \to 0} f(a+h) = f(a)$.

1. Start with the limit of the difference $f(a+h) - f(a)$:
2. $$\lim_{h \to 0} \left[f(a+h) - f(a)\right] = \lim_{h \to 0} \left(h \cdot \frac{f(a+h) - f(a)}{h}\right)$$
3. Use the product law for limits to split the expression:
4. $$= \lim_{h \to 0} h \cdot \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = 0 \cdot f'(a) = 0$$
5. This means $\lim_{h \to 0} [f(a+h) - f(a)] = 0$, which rearranges to $\
6. $\lim_{h \to 0} f(a+h) = f(a)$, the definition of continuity.

*Conclusion:* Therefore, differentiability at $x=a$ implies continuity at $x=a$. The contrapositive of this theorem is also useful: *If $f$ is not continuous at $x=a$, then $f$ cannot be differentiable at $x=a$*.

**Worked example:** The function $f(x)$ is defined as $f(x) = \begin{cases} x^2 + 3x & x \leq 2 \\ 7x + k & x > 2 \end{cases}$. Find the value of $k$ that makes $f$ differentiable at $x=2$.

1. By the core theorem, differentiability at $x=2$ requires continuity at $x=2$, so we first enforce the continuity condition.
2. Calculate the left-hand limit and value of $f$ at $x=2$:
3. $$f(2) = (2)^2 + 3(2) = 10 \implies \lim_{x \to 2^-} f(x) = 10$$
4. Calculate the right-hand limit:
5. $$\lim_{x \to 2^+} f(x) = 7(2) + k = 14 + k$$
6. Set the limits equal for continuity:
7. $$14 + k = 10 \implies k = -4$$
8. Confirm that one-sided derivatives match (required for full differentiability): Left derivative $f'_-(2) = 2(2) + 3 = 7$, right derivative $f'_+(2) = 7$, so derivatives match. Conclusion: $k=-4$ makes $f$ differentiable at $x=2$.

> **Exam tip:** If an AP FRQ asks for a constant to make a piecewise function differentiable, always solve for continuity first — AP graders award a point for this step, and it eliminates wrong answers in MCQ even if you don't finish the full problem.

## The Converse Is Not True: Continuity Does Not Imply Differentiability

The core theorem only guarantees that differentiability implies continuity; it does not work the other way around. A function can easily be continuous at a point but not differentiable at that point. This is the most commonly tested concept on the AP exam for this topic, as it probes whether students understand the difference between the two properties.

- **Corner (or Kink)**: The function is continuous, but the one-sided derivatives from the left and right are different finite values. No single well-defined slope exists.
- **Cusp**: The function is continuous, but the one-sided derivatives approach opposite infinities, so the slope is undefined.
- **Vertical Tangent**: The function is continuous, but the tangent line is vertical, so the slope is infinite (undefined), hence the derivative does not exist.

**Worked example:** Is $g(x) = |x^2 - 9|$ continuous and differentiable at $x=3$? Justify your answer.

1. First rewrite $g(x)$ by factoring: $x^2 - 9 = (x-3)(x+3)$, so $g(x)$ can be written piecewise as:
2. $$g(x) = \begin{cases} x^2 - 9 & x \geq 3 \\ 9 - x^2 & x < 3 \end{cases}$$
3. Check continuity: $\lim_{x \to 3^-} g(x) = 9 - 9 = 0$, $\lim_{x \to 3^+} g(x) = 9 - 9 = 0$, and $g(3)=0$, so $g$ is continuous at $x=3$.
4. Calculate left-hand derivative:
5. $$g'_-(3) = \lim_{h \to 0^-} \frac{g(3+h) - g(3)}{h} = \lim_{h \to 0^-} \frac{9 - (9 + 6h + h^2)}{h} = -6$$
6. Calculate right-hand derivative:
7. $$g'_+(3) = \lim_{h \to 0^+} \frac{(9 + 6h + h^2) - 9}{h} = 6$$
8. Since $g'_-(3) = -6 \neq 6 = g'_+(3)$, the two-sided derivative $g'(3)$ does not exist. Conclusion: $g$ is continuous at $x=3$ but not differentiable at $x=3$.

> **Exam tip:** Whenever you are asked to justify non-differentiability on an FRQ, you must explicitly compare one-sided derivatives or explain why the difference quotient limit is undefined — a conclusion without justification earns zero points.

## Identifying Non-Differentiable Points From a Graph

One of the most common MCQ questions on this topic asks you to count or identify non-differentiable points from the graph of a function. You can do this without any algebra by looking for four key features, in order:

- **Discontinuities**: Any jump, hole, infinite discontinuity, or mismatched point is automatically non-differentiable, by the contrapositive of the core theorem. Always count these first.
- **Corners/Kinks**: A sharp turn in a continuous graph, where the left and right slopes are visibly different.
- **Cusps**: A sharp point where the graph curves inward from both sides to meet at the point, with opposite infinite slopes.
- **Vertical Tangents**: A point where the graph becomes vertical, so the slope is infinite (undefined) even though the function is continuous.

**Worked example:** The graph of $h(x)$ on the interval $(-3, 5)$ has a jump discontinuity at $x=-2$, a sharp corner at $x=1$, a vertical tangent at $x=4$, and is smooth everywhere else. How many non-differentiable points are there on $(-3,5)$, and why?

1. First, count discontinuities: $x=-2$ is discontinuous, so it is non-differentiable.
2. Next, check continuous points for non-differentiability: $x=1$ is continuous, but has a sharp corner with unequal one-sided slopes, so it is non-differentiable.
3. $x=4$ is continuous, but has a vertical tangent, so the slope is undefined, hence non-differentiable.
4. All other points are smooth and continuous, so they are differentiable. Total non-differentiable points: 3, at $x=-2$, $x=1$, and $x=4$.

> **Exam tip:** If the question asks for how many points, double-check that you didn't miss discontinuities — most students lose points here by only counting sharp turns and forgetting jumps/holes.

## AP-Style Worked Practice Problems

**Worked example:** Let $f(x) = \begin{cases} \dfrac{x^2 - 3x}{x-3} & x \neq 3 \\ 3 & x = 3 \end{cases}$. Which of the following statements is true?<br>A) $f$ is not continuous at $x=3$, hence not differentiable at $x=3$.<br>B) $f$ is continuous at $x=3$, but not differentiable at $x=3$.<br>C) $f$ is continuous at $x=3$, and $f$ is differentiable at $x=3$, and $f'(3) = 1$.<br>D) $f$ is continuous at $x=3$, and $f$ is differentiable at $x=3$, and $f'(3) = 3$.

1. First, check continuity at $x=3$. For $x \neq 3$, factor the numerator: $\frac{x(x-3)}{x-3} = x$, so $\lim_{x \to 3} f(x) = 3$, which equals $f(3)=3$. This means $f$ is continuous at $x=3$, eliminating option A.
2. For all points around $x=3$, $f(x)=x$, a linear function with constant derivative $f'(x)=1$. There is no corner, discontinuity, or vertical tangent at $x=3$, so $f$ is differentiable at $x=3$ with derivative 1, eliminating options B and D. The correct answer is C.

**Worked example:** Let $f(x) = \begin{cases} mx + b & x < 2 \\ -x^2 + 8x - 7 & x \geq 2 \end{cases}$.<br>(a) Find values of $m$ and $b$ such that $f$ is differentiable at $x=2$. Justify your answer.<br>(b) Is $f$ differentiable at $x=0$? Justify your answer.<br>(c) How many points of non-differentiability does $f$ have for all real numbers $x$? Explain.

1. Part (a): For $f$ to be differentiable at $x=2$, it must first be continuous at $x=2$. Calculate $f(2) = -(2)^2 + 8(2) - 7 = 5$. The left-hand limit is $\
2. $\lim_{x \to 2^-} (mx + b) = 2m + b$, so first equation: $2m + b = 5$.
3. For differentiability, one-sided derivatives must be equal: right-hand derivative $f'_+(2) = -2(2) + 8 = 4$, left-hand derivative $f'_-(2) = m$, so $m=4$. Substitute back: $2(4) + b = 5 \implies b = -3$. These values satisfy both conditions.
4. Part (b): $x=0$ is an interior point of $(-\infty, 2)$, where $f(x) = 4x - 3$, a linear polynomial that is differentiable at all real points. Therefore, $f$ is differentiable at $x=0$.
5. Part (c): The only possible non-differentiable point is the piecewise boundary $x=2$. After choosing $m=4$ and $b=-3$, $f$ is differentiable at $x=2$, and all other points are in either the linear or quadratic domain, which are both differentiable everywhere. Therefore, $f$ has 0 points of non-differentiability.

**Worked example:** The position of a robot moving along a straight track is given by $s(t) = |t^2 - 6t + 8|$ for $0 \leq t \leq 5$, where $s(t)$ is distance from the start in meters, and $t$ is time in seconds. At which times $t$ in $(0,5)$ is the robot's instantaneous velocity undefined? Justify your answer, and interpret the result.

1. First, factor the quadratic inside the absolute value: $t^2 - 6t + 8 = (t-2)(t-4)$, so the expression changes sign at $t=2$ and $t=4$, both inside $(0,5)$.
2. $s(t)$ is continuous at both points: $\
3. $\lim_{t \to 2} s(t) = 0 = s(2)$ and $\
4. $\lim_{t \to 4} s(t) = 0 = s(4)$.
5. Check one-sided derivatives: at $t=2$, $s'_-(2) = -2$ and $s'_+(2) = 2$, so derivatives do not match. At $t=4$, $s'_-(4) = -2$ and $s'_+(4) = 2$, so derivatives also do not match here.
6. Instantaneous velocity is the derivative of position, so velocity is undefined at $t=2$ and $t=4$ seconds. In context, this means the robot reverses direction abruptly at these times, with no smooth transition between moving toward and away from the starting sensor.

## Common pitfalls

- **Wrong:** Assuming continuity at $x=a$ guarantees differentiability at $x=a$, and stopping after checking continuity.
  - Why it fails: Students remember the core theorem but forget it only works one direction, and the AP exam specifically tests this common misconception.
  - Correct: Always follow a continuity conclusion with a check of one-sided derivatives or the slope before concluding differentiability.
- **Wrong:** For piecewise functions with two unknown constants, matching one-sided derivatives first before enforcing continuity.
  - Why it fails: If the function is discontinuous, the derivative cannot exist even if slopes match, leading to wrong values for the constants.
  - Correct: Always solve for the constants that give continuity first, then solve for the second constant that matches derivatives.
- **Wrong:** Calling a vertical tangent point a discontinuity, and justifying non-differentiability with the wrong reason.
  - Why it fails: Vertical tangents always occur at continuous points, so students confuse undefined slope with a break.
  - Correct: For a vertical tangent at $x=a$, state that $f$ is continuous at $a$, but the derivative approaches $\pm\infty$, so the derivative does not exist.
- **Wrong:** Using the limit of $f'(x)$ as $x \to a$ instead of the difference quotient to find one-sided derivatives for piecewise functions.
  - Why it fails: Students assume that $\
- **Wrong:** 
  - Why it fails: $\lim_{x \to a} f'(x) = f'(a)$, but this is only true if $f'$ is continuous at $a$, which is not guaranteed.
  - Correct: Always use the difference quotient definition of one-sided derivatives when checking differentiability at a piecewise boundary to earn full justification credit.
- **Wrong:** Counting a hole (removable discontinuity) as a differentiable point because the limit of $f(x)$ exists at the point.
  - Why it fails: Students confuse the limit of $f$ with the value of $f$, so they incorrectly assume continuity holds.
  - Correct: Always confirm that $\
- **Wrong:** 
  - Correct: $\lim_{x \to a} f(x) = f(a)$ for continuity; if this fails, the point is discontinuous and non-differentiable.

## Cheatsheet

| Category | Rule/Result | Notes |
| --- | --- | --- |
| Core Theorem | If $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$ | Only works in this direction; never reverse this implication on the exam. |
| Contrapositive of Core Theorem | If $f$ is not continuous at $x=a$, then $f$ is not differentiable at $x=a$ | Use this to quickly rule out differentiability at any discontinuous point. |
| Key Result | Continuity does not imply differentiability | This is the most commonly tested misconception on the AP exam. |
| Differentiability Test | $f'(a)$ exists iff $f'_-(a) = f'_+(a)$ (both finite) | Always use this for piecewise functions at their boundary points. |
| Types of Non-Differentiable Points | 1. Discontinuity, 2. Corner/Kink, 3. Cusp, 4. Vertical Tangent | Only discontinuity is non-continuous; the other three occur at continuous points. |
| Standard Function Differentiability | All polynomials, exponentials, logs, and trig functions are differentiable at all domain points | Any interior point of the domain of these functions is automatically differentiable. |
| Step for Piecewise Constant Problems | 1. Enforce continuity at the split point, 2. Match one-sided derivatives | Do this order, not the reverse, to avoid mistakes. |

## What's next

Understanding the relationship between differentiability and continuity is a foundational prerequisite for all subsequent differentiation topics in AP Calculus AB. All basic differentiation rules for power, exponential, trigonometric, product, and quotient functions assume the function you are working with is differentiable at the point of interest, so failing to identify non-differentiable points will lead to incorrect derivative calculations. Mastery of this concept will also help you with later topics like graphing derivatives and identifying critical points for extrema problems.

- [AP Calculus AB Power rule](https://www.owlsprep.com/study/ap-calculus-ab-u2-power-rule/)
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