AP Calculus AB 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 AB.

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. Yes; continuity at $x=8$ implies $f'(8)$ exists.
2. No; a sharp point (corner/cusp) prevents differentiability at $x=8$. (correct answer)
3. Yes; sharp points imply the tangent line is horizontal.
4. No; sharp points imply a jump discontinuity.
5. Yes; $f'(8)$ exists because $f(8)$ is defined.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, analyzing how sharp points affect differentiability. The function is continuous at x=8, meeting the continuity requirement. However, a sharp point like a corner or cusp means the one-sided derivatives do not match or are undefined. Thus, no unique tangent line exists, so f'(8) does not exist. A tempting distractor might claim yes because of continuity, but additional conditions on slopes are needed. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 2

1. At both $x=0$ and $x=5$ because differentiability depends only on limits.
2. At neither $x=0$ nor $x=5$ because corners and discontinuities prevent differentiability. (correct answer)
3. Only at $x=0$ because corners are differentiable if continuous.
4. Only at $x=5$ because holes do not affect differentiability.
5. At both points because $f$ can still have tangent lines there.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, comparing corners and holes. A corner at x=0 means unequal one-sided derivatives, preventing differentiability despite continuity. A hole at x=5 causes discontinuity, which also blocks differentiability. Thus, f is differentiable at neither point. A tempting distractor might say at both because of possible tangent lines, but both features violate key conditions. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 3

1. Yes; no break means differentiable.
2. No; a cusp prevents a well-defined finite tangent slope. (correct answer)
3. Yes; cusps occur only when the derivative is $0$.
4. No; a cusp implies a jump discontinuity at $x=2$.
5. Yes; if $f(2)$ exists, then $f'(2)$ exists.

Explanation: This question explores differentiability and continuity at cusps without breaks. A cusp at x=2 with no break means the function is continuous, but the slopes approach infinity or mismatch, preventing a finite derivative. Thus, f'(2) does not exist due to the lack of a well-defined tangent. The absence of a break confirms continuity but not differentiability. Choice A is a tempting distractor, suggesting no break means differentiable, but it fails because cusps violate the finite slope requirement. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 4

1. Yes; both sides are polynomials/linear, so $f'(0)$ exists.
2. No; the one-sided derivatives at $0$ are different, so $f'(0)$ does not exist. (correct answer)
3. Yes; $f$ is continuous at $0$, so $f'(0)$ exists.
4. No; $f$ has a jump discontinuity at $0$.
5. Yes; a corner means the derivative equals the average slope.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, exploring piecewise functions and their derivatives at join points. For the given piecewise function, it is continuous at x=0 since both pieces approach 0. However, the left-hand derivative is 0 (from 2x at x=0-), while the right-hand derivative is -1 (from -x). This difference means the derivative does not exist at x=0. A tempting distractor might say yes because of continuity, but equal one-sided derivatives are also required. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 5

1. No; differentiability does not imply continuity.
2. Yes; differentiability at $x=7$ implies $f$ is continuous at $x=7$. (correct answer)
3. Yes; $f$ must have a corner at $x=7$.
4. No; $f$ must have a jump at $x=7$.
5. Yes; $f$ must have a vertical tangent at $x=7$.

Explanation: This question explores the implication of differentiability on continuity. If f is differentiable at x=7, then by theorem, it must be continuous there, as differentiability requires the limit of the difference quotient to exist, which implies continuity. The existence of f'(7) ensures the function approaches the same value from both sides at x=7. No exceptions exist where a function is differentiable but discontinuous. Choice A is a tempting distractor, claiming differentiability does not imply continuity, but it fails because the theorem states the opposite. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 6

1. Yes; piecewise linear functions are differentiable at all $x$.
2. No; the one-sided slopes differ at $x=2$, so $f'(2)$ does not exist. (correct answer)
3. Yes; the function is continuous at $x=2$, so $f'(2)$ exists.
4. No; $f$ has a jump discontinuity at $x=2$, so $f'(2)$ does not exist.
5. Yes; matching endpoints guarantee matching derivatives.

Explanation: This question probes differentiability versus continuity for piecewise linear functions. For this f(x), at x=2, the function is continuous since both pieces meet at 3, but the left slope is 1 and the right is 3. These differing one-sided slopes mean f'(2) does not exist. Continuity is present, but differentiability requires matching derivatives from both sides. Choice C is a tempting distractor, stating continuity ensures the derivative exists, but it fails because unequal slopes create a corner. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 7

1. Yes; $f$ is continuous at $1$, so $f'(1)$ exists.
2. No; $f$ has a vertical tangent at $x=1$, so the derivative does not exist. (correct answer)
3. Yes; square roots remove corners, making $f$ differentiable at $1$.
4. No; $f$ has a jump discontinuity at $1$.
5. Yes; vertical tangents correspond to derivative $0$.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, examining functions with absolute values and roots. For f(x) = |x-1|^{1/2}, it is continuous at x=1 with value 0. However, the left-hand derivative approaches -infinity, and the right-hand approaches +infinity, creating a vertical tangent. This infinite mismatch means f'(1) does not exist. A tempting distractor might say yes because of continuity, but finite matching slopes are required. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 8

1. Yes; continuity implies the derivative must exist.
2. No; a vertical tangent indicates infinite slope, so $f'(0)$ does not exist. (correct answer)
3. Yes; infinite slope is still a derivative value.
4. No; vertical tangents occur only at discontinuities.
5. Yes; if $f(0)$ is defined, then $f'(0)$ exists.

Explanation: This question tests differentiability versus continuity in cases of vertical tangents. A vertical tangent at x=0 means the slope is infinite, so the limit of the difference quotient does not exist finitely. Therefore, f'(0) does not exist, even if f is continuous there. Continuity is required but not sufficient when the derivative is infinite. Choice A is a tempting distractor, claiming continuity implies differentiability, but it fails because infinite slopes prevent it. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 9

1. Yes; the function is continuous at $0$, so $f'(0)$ exists.
2. No; the one-sided derivatives at $0$ are different, so $f'(0)$ does not exist. (correct answer)
3. Yes; the derivative exists because $f(0)$ is defined.
4. No; the function has a jump discontinuity at $0$.
5. Yes; different one-sided derivatives imply a vertical tangent.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, analyzing piecewise linear functions. For this f(x), it is continuous at x=0 with value 2 matching both limits. However, the left-hand derivative is 1, while the right-hand is -1, creating a corner. This difference means f'(0) does not exist. A tempting distractor might say yes because of continuity, but slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 10

1. It exists because $f$ is continuous at $x=-4$.
2. It does not exist because a cusp prevents a well-defined tangent slope. (correct answer)
3. It equals $0$ because cusps occur at local maxima.
4. It exists and is infinite, so $f'(-4)$ is defined.
5. It does not exist because cusps imply jump discontinuities.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, addressing cusps in graphs. A cusp at x=-4 typically means the one-sided derivatives approach infinity or mismatch in a way that prevents a finite limit. This lack of a well-defined tangent slope means f'(-4) does not exist. Even if continuous, the cusp disrupts differentiability. A tempting distractor might say it equals 0 at maxima, but cusps do not ensure that. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 11

1. Yes; meeting at $x=1$ guarantees equal one-sided slopes.
2. No; the one-sided derivatives at $x=1$ are not equal, so $f'(1)$ does not exist. (correct answer)
3. Yes; piecewise functions are differentiable at the join point.
4. No; the function must be discontinuous at $x=1$.
5. Yes; continuity implies differentiability at $x=1$.

Explanation: This question examines differentiability and continuity for piecewise functions that meet at a point. For this f(x), at x=1, it is continuous since both pieces equal 1, but the left derivative is -1 and the right is 2. These unequal one-sided derivatives mean f'(1) does not exist. Continuity is necessary but insufficient without matching slopes. Choice E is a tempting distractor, claiming continuity implies differentiability, but it fails because of the derivative mismatch. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 12

1. Yes; vertical asymptotes imply a vertical tangent, so $f'(3)$ exists.
2. No; $f$ is not defined/continuous at $x=3$, so $f'(3)$ does not exist. (correct answer)
3. Yes; derivatives can exist even when functions are undefined.
4. No; vertical asymptotes create corners, so the derivative does not exist.
5. Yes; the left- and right-hand limits match at $x=3$.

Explanation: This question probes differentiability versus continuity near vertical asymptotes. A vertical asymptote at x=3 means f(x) approaches infinity or is undefined there, so f is not continuous at x=3. Without continuity, f'(3) cannot exist, as differentiability requires the function to be defined and continuous first. The asymptote prevents a finite limit. Choice A is a tempting distractor, suggesting asymptotes imply vertical tangents with existing derivatives, but it fails because asymptotes indicate discontinuity. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 13

1. Yes; smooth curves are differentiable even with holes.
2. No; a hole means $f$ is discontinuous at $1$, so $f'(1)$ does not exist. (correct answer)
3. Yes; if the limit exists, then the derivative exists.
4. No; a hole implies a cusp at $x=1$.
5. Yes; discontinuity implies a vertical tangent, so $f'(1)$ exists.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, investigating the impact of holes in graphs. The hole at x=1 means f(1) is undefined, resulting in a discontinuity. Differentiability requires continuity, so without it, f'(1) cannot exist. Even if the surrounding curve is smooth, the gap prevents a derivative at that point. A tempting distractor might say yes because of smoothness around the hole, but continuity is essential. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 14

1. Yes; vertical tangents mean the derivative is large but defined.
2. No; a vertical tangent indicates an infinite slope, so $f'(-1)$ does not exist. (correct answer)
3. Yes; if $f$ is continuous, then $f'(-1)$ exists.
4. No; vertical tangents imply a jump discontinuity.
5. Yes; vertical tangents imply a corner with equal one-sided slopes.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, exploring vertical tangents. A vertical tangent at x=-1 means the slope approaches infinity, which is not a finite value. Even if continuous, an infinite derivative is not considered to exist in the standard sense. Thus, f is not differentiable at x=-1. A tempting distractor might say yes because the derivative is 'large but defined,' but infinity is not a real number. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 15

1. Yes; jump discontinuities do not affect differentiability.
2. No; differentiability requires continuity, so $f'(1)$ does not exist. (correct answer)
3. Yes; if both one-sided slopes exist, then $f'(1)$ exists.
4. No; there must be a corner at $x=1$, so $f'(1)$ does not exist.
5. Yes; the derivative exists because $f(1)$ is defined.

Explanation: This question probes the relationship between differentiability and continuity, focusing on jump discontinuities. A jump discontinuity at x=1 means the left- and right-hand limits of f(x) differ, so f is not continuous there. Differentiability requires continuity as a prerequisite, so without it, f'(1) cannot exist. Even if one-sided slopes exist, the lack of continuity prevents the derivative from being defined. Choice A is a tempting distractor, suggesting jump discontinuities do not affect differentiability, but it fails because continuity is essential for differentiability. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 16

1. Yes; continuity at $x=-2$ guarantees differentiability.
2. No; a corner prevents differentiability even if $r$ is continuous. (correct answer)
3. Yes; corners occur only when the derivative equals $0$.
4. No; continuity implies a vertical tangent, so the derivative does not exist.
5. Yes; equal function values from both sides imply equal slopes.

Explanation: This question explores differentiability and continuity, highlighting corners in continuous functions. At x=-2, the corner means the left- and right-hand derivatives differ, so the overall derivative does not exist. Even though r is continuous, the mismatch in one-sided slopes prevents a unique tangent line. Continuity is necessary but not sufficient for differentiability in this case. Choice A is a tempting distractor, claiming continuity guarantees differentiability, but it fails because corners provide a counterexample where slopes do not match. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 17

1. Yes; smoothness on each side implies $f'(0)$ exists.
2. No; a jump discontinuity means $f$ is not continuous, so $f'(0)$ does not exist. (correct answer)
3. Yes; if one-sided derivatives exist, the derivative exists.
4. No; a jump indicates a corner, so $f'(0)$ does not exist.
5. Yes; discontinuities imply vertical tangents with defined slope.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, considering jump discontinuities. A jump at x=0 means the left- and right-hand limits differ, causing discontinuity. Since differentiability requires continuity, f'(0) cannot exist. The smoothness on each side is irrelevant without overall continuity. A tempting distractor might say yes if one-sided derivatives exist, but continuity is prerequisite. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 18

1. No; corners have unequal one-sided derivatives, so $m'(-1)$ does not exist. (correct answer)
2. Yes; sharp points mean the derivative is very large but exists.
3. Yes; if $m$ is defined at $x=-1$, then $m'(-1)$ exists.
4. No; corners imply a jump discontinuity at $x=-1$.
5. Yes; continuity at $x=-1$ guarantees differentiability.

Explanation: This question probes differentiability and continuity, specifically for sharp points or corners. A sharp point at x=-1 means the left- and right-hand derivatives are unequal, so m'(-1) does not exist. The function may be continuous, but the corner prevents a unique derivative. Sharp points typically indicate a change in slope direction or magnitude. Choice E is a tempting distractor, claiming continuity guarantees differentiability, but it fails because unequal one-sided derivatives violate the condition. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 19

1. No; $f$ is discontinuous at $x=0$, so $f'(0)$ does not exist. (correct answer)
2. Yes; both pieces are differentiable, so $f'(0)$ exists.
3. Yes; $f$ is continuous at $x=0$, so $f'(0)$ exists.
4. No; there is a corner at $x=0$, so $f'(0)$ does not exist.
5. Yes; the right-hand slope is $0$, so $f'(0)$ exists.

Explanation: This question evaluates the connection between differentiability and continuity for piecewise functions. For this f(x), at x=0, the left-hand limit is 0 but f(0)=1, creating a discontinuity. Since differentiability implies continuity, the lack of continuity means f'(0) does not exist. The pieces may be differentiable individually, but the jump at x=0 breaks continuity. Choice C is a tempting distractor, claiming continuity at x=0 ensures the derivative exists, but it fails because the function is actually discontinuous there. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 20

1. Yes; $\sin x$ is differentiable, so $f'(0)$ exists.
2. No; $f$ is discontinuous at $0$, so $f'(0)$ does not exist. (correct answer)
3. Yes; redefining one point never affects differentiability.
4. No; there is a corner at $0$, so $f'(0)$ does not exist.
5. Yes; discontinuity implies a vertical tangent, so $f'(0)$ exists.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, considering redefined points in otherwise continuous functions. For this f(x), lim x→0 sin x = 0, but f(0)=1, creating a discontinuity. Differentiability requires continuity, so f'(0) cannot exist here. The redefinition breaks the necessary limit agreement. A tempting distractor might say yes because redefining one point doesn't affect differentiability, but it does if it causes discontinuity. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.