Connecting differentiability and continuity — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The theorem that differentiability implies continuity, the false converse that continuity does not imply differentiability, testing differentiability at a point, identifying non-differentiable points, and working with piecewise functions.

You should already know: Limit definition of the derivative at a point. Definition of continuity at a point. Evaluating one-sided limits.

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 AB 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?

Connecting differentiability and continuity describes the logical relationship between two fundamental properties of functions, and is a core conceptual topic in AP Calculus AB Unit 2: Differentiation: Definition and Fundamental Properties, which contributes 10-12% of the total AP exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often as a standalone MCQ or as the first justification step in a multi-part FRQ. Differentiability at a point $x=a$ means the derivative $f^{\prime }(a)$ exists, i.e., the function has a well-defined tangent line with finite slope at that point. Continuity at $x=a$ means the limit of $f(x)$ as $x$ approaches $a$ equals $f(a)$, with no breaks, jumps, or holes at that point. The main goal of this topic is to establish how these two properties relate: which guarantees the other, and when a function can have one property without the other. Conceptual understanding of this relationship is frequently tested because it reveals whether students understand what it means for a derivative to exist, beyond just memorizing differentiation rules.

2. The Core Theorem: Differentiability Implies Continuity

The first and most important result for this topic is the core 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. If $f$ is differentiable at $a$, the derivative exists as a finite limit: $$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$ We can rearrange this to prove continuity, which requires ${\lim }_{h\to 0}f(a+h)=f(a)$. Start with the limit of the difference $f(a+h)-f(a)$: $$ \begin{align*} \lim_{h \to 0} \left[f(a+h) - f(a)\right] &= \lim_{h \to 0} \left(h \cdot \frac{f(a+h) - f(a)}{h}\right) \ &= \lim_{h \to 0} h \cdot \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \ &= 0 \cdot f'(a) = 0 \end{align*} $$ This means ${\lim }_{h\to 0}f(a+h)=f(a)$, which is exactly the definition of continuity at $a$. Intuitively, this makes sense: you cannot have a well-defined, finite-slope tangent line at a point where the function has a break, jump, or hole. The contrapositive of this theorem is also extremely useful: If $f$ is not continuous at $x=a$, then it cannot be differentiable at $x=a$. This lets us immediately rule out differentiability at any discontinuous point without extra work.

Worked Example

The function $f(x)$ is defined as $f(x)=\{\begin{array}{ll}{x}^2+3x& x\le 2\\ 7x+k& x>2\end{array}$. 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 $x=2$: $f(2)=(2{)}^2+3(2)=10$, so ${\lim }_{x\to 2^{-}}f(x)=10$.
3. Calculate the right-hand limit: ${\lim }_{x\to 2^{+}}f(x)=7(2)+k=14+k$.
4. Set the limits equal for continuity: $14+k=10⟹k=-4$.
5. Confirm that one-sided derivatives match (required for full differentiability): Left derivative $f_{-}^{\prime }(2)=2(2)+3=7$, right derivative $f_{+}^{\prime }(2)=7$, so derivatives match. The value $k=-4$ 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.

3. 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.

There are three common cases where a continuous function is non-differentiable at a point:

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

All of these cases are continuous at the point, but non-differentiable, which proves continuity does not guarantee differentiability.

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 $g(x)$ changes sign at $x=3$, and can be written piecewise as $g(x)=\{\begin{array}{ll}{x}^2-9& x\ge 3\\ 9-x^2& x<3\end{array}$.
2. 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 $x=3$.
3. Calculate left-hand derivative: $g_{-}^{\prime }(3)={\lim }_{h\to 0^{-}}\frac{g(3+h)-g(3)}{h}={\lim }_{h\to 0^{-}}\frac{9-(9+6h+h^2)}{h}={\lim }_{h\to 0^{-}}(-6-h)=-6$.
4. Calculate right-hand derivative: $g_{+}^{\prime }(3)={\lim }_{h\to 0^{+}}\frac{(9+6h+h^2)-9}{h}={\lim }_{h\to 0^{+}}(6+h)=6$.
5. Since $g_{-}^{\prime }(3)=-6\ne 6=g_{+}^{\prime }(3)$, the two-sided derivative $g^{\prime }(3)$ 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.

4. 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:

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

Any smooth portion of the graph that connects without breaks is differentiable everywhere in that portion.

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.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Let $f(x)=\{\begin{array}{ll}\frac{x^2-3x}{x-3}& x\ne 3\\ 3& x=3\end{array}$. Which of the following statements is true? A) $f$ is not continuous at $x=3$, hence not differentiable at $x=3$. B) $f$ is continuous at $x=3$, but not differentiable at $x=3$. C) $f$ is continuous at $x=3$, and $f$ is differentiable at $x=3$, and $f^{\prime }(3)=1$. D) $f$ is continuous at $x=3$, and $f$ is differentiable at $x=3$, and $f^{\prime }(3)=3$.

Worked Solution: First, check continuity at $x=3$. For $x\ne 3$, we factor the numerator to get $\frac{x(x-3)}{x-3}=x$, so ${\lim }_{x\to 3}f(x)={\lim }_{x\to 3}x=3$, which equals the defined value $f(3)=3$. This means $f$ is continuous at $x=3$, eliminating option A. For all points around $x=3$, $f(x)=x$, a linear function with constant derivative $f^{\prime }(x)=1$ for all $x$. There is no corner, discontinuity, or vertical tangent at $x=3$, so $f$ is differentiable at $x=3$ with derivative 1. This eliminates options B and D. The correct answer is C.

Question 2 (Free Response)

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

Worked Solution: (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 ${\lim }_{x\to 2^{-}}(mx+b)=2m+b$, so we get the first equation: $2m+b=5$. For differentiability, one-sided derivatives must be equal: the right-hand derivative is $f_{+}^{\prime }(2)=-2(2)+8=4$, and the left-hand derivative is $f_{-}^{\prime }(2)=m$, so $m=4$. Substitute back to get $2(4)+b=5⟹b=-3$. These values satisfy both continuity and equal derivatives, so $m=4$, $b=-3$. (b) $x=0$ is an interior point of the interval $(-\infty ,2)$, where $f(x)=4x-3$, a linear (polynomial) function that is differentiable at all real points. Therefore, $f$ is differentiable at $x=0$. (c) The only possible point of non-differentiability is at the piecewise boundary $x=2$. After choosing $m=4$ and $b=-3$, $f$ is differentiable at $x=2$, and all other points are either in the linear domain (all differentiable) or the quadratic domain (all polynomials are differentiable everywhere). Therefore, $f$ has 0 points of non-differentiability for all real $x$.

Question 3 (Application / Real-World Style)

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

Worked Solution: 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 $(0,5)$. $s(t)$ is continuous at both points, since ${\lim }_{t\to 2}s(t)=0=s(2)$ and ${\lim }_{t\to 4}s(t)=0=s(4)$. Next, check one-sided derivatives: at $t=2$, $s_{-}^{\prime }(2)=2(2)-6=-2$ and $s_{+}^{\prime }(2)=-2(2)+6=2$, so the derivatives do not match. At $t=4$, $s_{-}^{\prime }(4)=-2(4)+6=-2$ and $s_{+}^{\prime }(4)=2(4)-6=2$, so derivatives do not match here either. All other points are in intervals where $s(t)$ is quadratic, hence differentiable. Instantaneous velocity is the derivative of position, so velocity is undefined at $t=2$ seconds 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.

7. Quick Reference Cheatsheet

8. What's Next

Understanding the relationship between differentiability and continuity is a foundational prerequisite for all subsequent differentiation topics in AP Calculus AB. Immediately after this topic, you will learn the basic differentiation rules for power, exponential, trigonometric, product, and quotient functions — all of these rules assume the function you are differentiating is differentiable at the point of interest, so if you fail to identify non-differentiable points, you will incorrectly compute derivatives where they do not exist. This topic also feeds into later heavily tested concepts such as the Mean Value Theorem (which requires differentiability to apply), finding critical points, and sketching graphs of derivatives. Without this foundation, you will struggle to correctly work with piecewise or absolute value functions, which are common in AP FRQ sections.

• Derivative rules: power rule
• Derivative rules: product and quotient
• The Mean Value Theorem
• Graphing the derivative of a function