AP Calculus BC Quiz: Connecting Differentiability And Continuity

What this quiz covers

This quiz focuses on Connecting Differentiability And Continuity, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. If $h$ is continuous at $x=0$, then $h$ is differentiable at $x=0$.
2. If $h$ is differentiable at $x=0$, then $h$ is continuous at $x=0$. (correct answer)
3. If $h$ is not differentiable at $x=0$, then $h$ is not continuous at $x=0$.
4. If $h$ is continuous at $x=0$, then $h$ is not continuous at $x=0$.
5. If $h$ is not continuous at $x=0$, then $h$ is differentiable at $x=0$.

Explanation: This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because the finite derivative limit ensures the function's limit matches its value at that point. Specifically, for h(x)=|x|, if differentiable at x=0, the difference quotient limit would require continuity, but here we use the theorem in reverse. However, continuity does not imply differentiability, as seen in the absolute value function where left and right slopes differ. A tempting distractor is choice A, which states that if h is continuous at x=0, then it is differentiable there, but this fails because the corner at x=0 means the derivative does not exist despite continuity. A transferable implication strategy is to test one-sided limits of the difference quotient to check differentiability after confirming continuity.

Question 2

1. $f$ is differentiable at $x=2$ and continuous at $x=2$.
2. $f$ is continuous at $x=2$ but not differentiable at $x=2$. (correct answer)
3. $f$ is differentiable at $x=2$ but not continuous at $x=2$.
4. $f$ is neither continuous at $x=2$ nor differentiable at $x=2$.
5. Nothing can be concluded about continuity at $x=2$ from the given information.

Explanation: This question tests the connection between differentiability and continuity. Since f(2) = 5 and lim_{x→2} f(x) = 5, we have continuity at x = 2 because the function value equals the limit. However, we're told that f'(2) does not exist, meaning f is not differentiable at x = 2. This demonstrates that continuity does not guarantee differentiability—a function can be continuous at a point but fail to have a derivative there (like at a corner or cusp). Choice C is impossible because differentiability always implies continuity. When a function is continuous but the derivative doesn't exist, look for geometric features like corners, cusps, or vertical tangents.

Question 3

1. $u'(0)$ must exist because $u$ is differentiable nearby.
2. $u'(0)$ must not exist because differentiability at $0$ would imply continuity at $0$. (correct answer)
3. $u'(0)$ must exist if and only if $u$ is bounded near $0$.
4. $u'(0)$ must not exist only if $u$ is continuous at $0$.
5. $u'(0)$ must exist because continuity is not required for differentiability.

Explanation: This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice E, which fails because it claims differentiability without continuity, which contradicts the theorem. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

Question 4

1. If $s$ is continuous at $x=5$, then $s$ is differentiable at $x=5$.
2. If $s$ is differentiable at $x=5$, then $s$ is continuous at $x=5$. (correct answer)
3. If $s$ is not differentiable at $x=5$, then $s$ is not continuous at $x=5$.
4. If $s$ is continuous at $x=5$, then $s$ is not continuous at $x=5$.
5. If $s$ is not continuous at $x=5$, then $s$ is differentiable at $x=5$.

Explanation: This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because a finite derivative requires the function to limit to its value. Specifically, a vertical tangent means the derivative limit is infinite, so not differentiable, but continuity holds. However, continuity does not imply differentiability, as vertical tangents demonstrate. A tempting distractor is choice A, which states that if s is continuous at x=5, then it is differentiable there, but this fails because the vertical tangent prevents a finite derivative despite continuity. A transferable implication strategy is to examine the behavior of slopes near the point to distinguish continuity from differentiability.

Question 5

1. If $g$ is continuous at $x=0$, then $g$ is differentiable at $x=0$.
2. If $g$ is differentiable at $x=0$, then $g$ is not continuous at $x=0$.
3. If $g$ is differentiable at $x=0$, then $g$ is continuous at $x=0$. (correct answer)
4. If $g$ is not differentiable at $x=0$, then $g$ is not continuous at $x=0$.
5. If $g$ is continuous at $x=0$, then $g$ is not continuous at $x=0$.

Explanation: This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because the derivative's existence ensures the limit of the function equals its value, derived from the difference quotient limit. Specifically, the limit of [g(x) - g(0)]/(x-0) existing implies lim g(x) = g(0). However, continuity does not imply differentiability, as sharp corners allow continuity but prevent a unique tangent slope. A tempting distractor is choice A, which states that if g is continuous at x=0, then it is differentiable there, but this fails because the sharp corner means unequal left and right derivatives despite continuity and no hole. A transferable implication strategy is to recall that differentiability is stronger than continuity, so use graphical features like corners to identify where continuity holds but differentiability does not.

Question 6

1. $f$ is continuous at $x=1$ but not differentiable at $x=1$. (correct answer)
2. $f$ is differentiable at $x=1$ but not continuous at $x=1$.
3. $f$ is both differentiable and continuous at $x=1$.
4. $f$ is neither differentiable nor continuous at $x=1$.
5. Unequal one-sided derivatives imply $f$ is discontinuous at $x=1$.

Explanation: This question directly addresses what unequal one-sided derivatives mean for differentiability and continuity. If left and right derivatives at x = 1 exist but are unequal, then f'(1) doesn't exist (the derivative requires both one-sided derivatives to exist and be equal). However, the problem states f is continuous at x = 1, which is perfectly possible—think of a corner or cusp where the graph doesn't break but changes direction sharply. This situation (continuous but not differentiable) is common at corners, cusps, and points with vertical tangents. Choice E incorrectly claims unequal derivatives imply discontinuity, but corners are continuous. When one-sided derivatives differ, you have non-differentiability, but continuity is a separate question determined by limits and function values.

Question 7

1. If $r$ is not continuous at $x=a$, then $r$ is differentiable at $x=a$.
2. If $r$ is continuous at $x=a$, then $r$ is differentiable at $x=a$.
3. If $r$ is differentiable at $x=a$, then $r$ is continuous at $x=a$. (correct answer)
4. If $r$ is not differentiable at $x=a$, then $r$ is not continuous at $x=a$.
5. If $r$ is continuous at $x=a$, then $r$ is not differentiable at $x=a$.

Explanation: This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because the existence of the limit of the difference quotient ensures the function is continuous. Specifically, if r is differentiable at x=a, then by theorem, it must be continuous at x=a. However, continuity does not imply differentiability, as many continuous functions lack derivatives at certain points. A tempting distractor is choice B, which states that if r is continuous at x=a, then it is differentiable there, but this fails because continuity is necessary but not sufficient for differentiability. A transferable implication strategy is to apply the contrapositive: if a function is not continuous, it cannot be differentiable, helping rule out options quickly.

Question 8

1. $f$ must be differentiable at $x=1$.
2. $f$ must be continuous at $x=1$. (correct answer)
3. $f$ must be differentiable at $x=1$ and continuous at $x=1$.
4. $f$ must be neither differentiable at $x=1$ nor continuous at $x=1$.
5. $f$ must be differentiable at $x=1$ but may fail to be continuous at $x=1$.

Explanation: This question tests understanding of boundary behavior for differentiability and continuity. The problem states f is continuous on the closed interval [1,3], which explicitly includes continuity at the endpoints x = 1 and x = 3. However, f is only differentiable on the open interval (1,3), which excludes the endpoints—we know nothing about differentiability at x = 1. This is a common scenario in calculus theorems like the Mean Value Theorem, where continuity extends to endpoints but differentiability might not. Choice A incorrectly assumes differentiability at x = 1, but the given information doesn't guarantee this. When dealing with closed versus open intervals, pay attention to what's guaranteed at endpoints—continuity often extends there, but differentiability might not.

Question 9

1. $f$ is continuous at $x=7$ but not differentiable at $x=7$.
2. $f$ is differentiable at $x=7$ but not continuous at $x=7$.
3. $f$ is neither continuous at $x=7$ nor differentiable at $x=7$. (correct answer)
4. $f$ is both continuous and differentiable at $x=7$.
5. A jump discontinuity guarantees $f'(7)$ exists.

Explanation: This question tests the relationship between discontinuity types and differentiability. A jump discontinuity at x = 7 means the left and right limits exist but are unequal, so lim_{x→7} f(x) doesn't exist, making f discontinuous at x = 7. Since differentiability requires continuity as a prerequisite, f cannot be differentiable at x = 7 either. The function fails both conditions—it has a break (discontinuous) and certainly no tangent line can exist across a jump. Choice E is absurd because discontinuities prevent derivatives from existing. Remember the hierarchy: for differentiability, you must first have continuity; any discontinuity automatically rules out differentiability.

Question 10

1. $t'(1)$ exists because $t(1)$ is defined.
2. $t'(1)$ exists only if $t$ is continuous at $x=1$.
3. $t'(1)$ does not exist because differentiability implies continuity. (correct answer)
4. $t'(1)$ exists if and only if \lim_{x\to1}t(x)=0.
5. $t'(1)$ exists because continuity is sufficient but not necessary.

Explanation: This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice A, which fails because the derivative requires more than just the function being defined; continuity is necessary. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

Question 11

1. $p$ is differentiable at $x=1$ and therefore continuous at $x=1$.
2. $p$ is continuous at $x=1$ but not differentiable at $x=1$. (correct answer)
3. $p$ is differentiable at $x=1$ but not continuous at $x=1$.
4. $p$ is neither continuous nor differentiable at $x=1$.
5. $p$ is continuous at $x=1$ and therefore differentiable at $x=1$.

Explanation: This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice C, which fails because it suggests differentiability without continuity, which is impossible. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

Question 12

1. $q$ is differentiable at $x=-2$ because it is continuous there.
2. $q$ is not continuous at $x=-2$ because it is not differentiable there.
3. $q$ is continuous at $x=-2$ but not differentiable at $x=-2$. (correct answer)
4. $q$ is differentiable at $x=-2$ if and only if $q(-2)=0$.
5. $q$ is neither continuous nor differentiable at $x=-2$.

Explanation: This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice A, which fails because continuity does not imply differentiability, as stated in the question where the derivative does not exist. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

Question 13

1. $h$ is continuous at $x=3$. (correct answer)
2. $h$ is discontinuous at $x=3$.
3. $h$ has a corner at $x=3$.
4. $h$ has a vertical tangent at $x=3$.
5. $h$ is continuous at $x=3$ if and only if $h'(3)=0$.

Explanation: This question tests the connection between differentiability and continuity, a key concept in calculus. Differentiability at a point requires that the function has a well-defined tangent line, which implies the function must be continuous there because the limit of the difference quotient must exist and match the function's behavior. However, continuity alone does not guarantee differentiability, as a function can be continuous but have a sharp corner or cusp where no tangent exists. For example, the absolute value function is continuous everywhere but not differentiable at the vertex. A tempting distractor is choice B, which fails because differentiability ensures continuity, so the function cannot be discontinuous if differentiable. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.

Question 14

1. $f$ is continuous at $x=4$ but not differentiable at $x=4$. (correct answer)
2. $f$ is differentiable at $x=4$ but not continuous at $x=4$.
3. $f$ is both continuous and differentiable at $x=4$.
4. $f$ is neither continuous nor differentiable at $x=4$.
5. Continuity at $x=4$ implies differentiability at $x=4$.

Explanation: This question connects geometric features to differentiability and continuity. A corner at x = 4 means the graph has a sharp turn where left and right derivatives exist but are unequal, making f non-differentiable at x = 4. However, since lim_{x→4} f(x) = f(4), the function is continuous at x = 4—there's no break or jump, just a sharp turn. This exemplifies how continuity doesn't guarantee differentiability: corners are continuous but not smooth enough for a unique tangent line. Choice E incorrectly reverses the implication—continuity never implies differentiability. When you see geometric terms like "corner," "cusp," or "vertical tangent," the function is typically continuous but not differentiable at that point.

Question 15

1. $f$ is differentiable at $x=3$ but not continuous at $x=3$.
2. $f$ is continuous at $x=3$ but not differentiable at $x=3$. (correct answer)
3. $f$ is both continuous and differentiable at $x=3$.
4. $f$ is neither continuous nor differentiable at $x=3$.
5. $f$ is differentiable at $x=3$ because absolute value functions are always differentiable.

Explanation: This question examines differentiability at points where absolute value creates corners. The function f(x) = |x - 3| equals (x - 3) for x ≥ 3 and -(x - 3) for x < 3, creating a V-shaped graph with a corner at x = 3. At this corner, f is continuous (no break in the graph) since lim_{x→3} f(x) = 0 = f(3). However, the left derivative is -1 and the right derivative is +1, so f'(3) doesn't exist due to the sharp turn. Choice E incorrectly claims absolute value functions are always differentiable, ignoring corners. When analyzing piecewise or absolute value functions, check points where the formula changes—these often create corners that are continuous but not differentiable.

Question 16

1. $p$ is differentiable at $x=4$ but not continuous at $x=4$.
2. $p$ is continuous at $x=4$ but not differentiable at $x=4$.
3. $p$ is not differentiable at $x=4$. (correct answer)
4. $p$ is differentiable at $x=4$ if the one-sided derivatives match.
5. $p$ is differentiable at $x=4$ because discontinuities do not affect derivatives.

Explanation: This question examines how discontinuity affects differentiability. A jump discontinuity at x = 4 means that lim_{x→4^-} p(x) ≠ lim_{x→4^+} p(x), creating a sudden jump in the function's values. For a function to be differentiable at a point, it must first be continuous there—this is a necessary condition. Since p has a jump discontinuity at x = 4, it fails the continuity requirement and therefore cannot possibly be differentiable at that point. Choice E represents a dangerous misconception, suggesting that discontinuities don't affect derivatives, when in fact discontinuity automatically prevents differentiability. When a function has any type of discontinuity at a point, immediately conclude it cannot be differentiable there.

Question 17

1. $t$ is continuous at $x=1$. (correct answer)
2. $t$ is discontinuous at $x=1$.
3. $t$ is continuous at $x=1$ only if $t(1)=7$.
4. $t$ is differentiable at $x=1$ only if $\lim_{x\to1}t(x)$ does not exist.
5. $t$ is differentiable at $x=1$ but may fail to be continuous there.

Explanation: This question directly applies the theorem that differentiability implies continuity. If t'(1) exists and equals 7, then t must be differentiable at x = 1. Since differentiability at a point automatically guarantees continuity at that point, we can conclude that t is continuous at x = 1. The specific value of the derivative (7 in this case) is irrelevant to the continuity conclusion—any finite derivative value ensures continuity. Choice C incorrectly suggests that continuity depends on a specific function value, while choice E impossibly claims differentiability without continuity. Remember: the existence of a derivative at a point is sufficient to guarantee continuity at that point.

Question 18

1. $v$ is continuous at $x=2$ but not necessarily differentiable at $x=2$.
2. $v$ is differentiable at $x=2$ and therefore continuous at $x=2$. (correct answer)
3. $v$ is continuous at $x=2$ only if $v'(2)=0$.
4. $v$ is differentiable at $x=2$ only if it is discontinuous at $x=2$.
5. $v$ is neither continuous nor differentiable at $x=2$.

Explanation: This question combines interval properties with the differentiability-continuity relationship. If v is differentiable on the open interval (1,3), then v is differentiable at every point in that interval, including x = 2. Since differentiability at a point implies continuity at that point, v must be continuous at x = 2. The fact that v is also continuous on the closed interval [1,3] provides additional information but doesn't change our conclusion about x = 2. Choice A suggests uncertainty about differentiability when we're explicitly told v is differentiable throughout (1,3). Choice C incorrectly ties continuity to a specific derivative value. When a function is differentiable on an open interval containing a point, it's both differentiable and continuous at that point.

Question 19

1. $u$ is continuous and differentiable at $x=0$.
2. $u$ is continuous but not differentiable at $x=0$.
3. $u$ is differentiable but not continuous at $x=0$.
4. $u$ is neither continuous nor differentiable at $x=0$. (correct answer)
5. $u$ is differentiable at $x=0$ because $\sin x$ is differentiable.

Explanation: This question tests understanding of point discontinuities and their effect on differentiability. For the function u, we have lim_{x→0} sin(x) = sin(0) = 0, but u(0) = 2. Since the limit (0) doesn't equal the function value (2), u has a removable discontinuity at x = 0 and is therefore not continuous there. Because differentiability requires continuity as a prerequisite, u cannot be differentiable at x = 0 either. Choice E incorrectly assumes that the differentiability of sin(x) for x ≠ 0 somehow extends to x = 0, ignoring the discontinuity created by the special value u(0) = 2. When a piecewise function assigns a special value that doesn't match the limit, it creates a discontinuity that prevents both continuity and differentiability.

Question 20

1. If $u$ is continuous at $x=3$, then $u$ is differentiable at $x=3$.
2. If $u$ is differentiable at $x=3$, then $u$ is continuous at $x=3$. (correct answer)
3. If $u$ is not differentiable at $x=3$, then $u$ is not continuous at $x=3$.
4. If $u$ is continuous at $x=3$, then $u$ is not continuous at $x=3$.
5. If $u$ is not continuous at $x=3$, then $u$ is differentiable at $x=3$.

Explanation: This question tests the connection between differentiability and continuity of a function at a point. Differentiability at a point implies continuity there because without continuity, the difference quotient cannot have a limit. Specifically, a jump discontinuity prevents both continuity and differentiability. However, continuity does not imply differentiability, though here neither holds. A tempting distractor is choice A, which states that if u is continuous at x=3, then it is differentiable there, but this fails because even if continuous, differentiability is not guaranteed, and here it's not continuous anyway. A transferable implication strategy is to use the theorem's contrapositive: non-continuity implies non-differentiability, simplifying analysis of discontinuous functions.