Connecting Differentiability and Continuity - AP Calculus AB

1. The graph of $f$ has a corner at $x=3$; which statement must be true about $f'(3)$?
   Answer: It does not exist because the one-sided derivatives are not equal at $x=3$. This question assesses the connection between differentiability and continuity in AP Calculus AB, focusing on corners and their effect on derivatives. A corner at x=3 implies the function is continuous but has unequal one-sided derivatives. This inequality prevents the existence of a single derivative value. Therefore, f'(3) does not exist. A tempting distractor might say it equals 0 at local extrema, but corners do not guarantee that and still lack a derivative. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.
2. The graph of $f$ has a cusp at $x=2$ but no break; does $f'(2)$ exist?
   Answer: No; a cusp prevents a well-defined finite tangent slope. 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.
3. For $f(x)=\begin{cases}x^2,&x\le0\\-x,&x>0\end{cases}$, does $f'(0)$ exist, and why?
   Answer: No; the one-sided derivatives at $0$ are different, so $f'(0)$ does not exist. 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.
4. For $f(x)=\begin{cases}x^3,&x<1\\x^3+2,&x\ge1\end{cases}$, does $f'(1)$ exist, and why?
   Answer: No; there is a jump discontinuity at $x=1$, so $f'(1)$ does not exist. This question tests differentiability versus continuity in piecewise functions with jumps. For this f(x), at x=1, the left-hand limit is 1 and the right-hand is 3, creating a jump discontinuity. Without continuity, f'(1) cannot exist, regardless of the pieces being differentiable elsewhere. The added constant causes the discontinuity. Choice C is a tempting distractor, claiming a jump still allows a derivative, but it fails because continuity is required. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.
5. A graph of $f$ shows a vertical tangent at $x=-1$; is $f$ differentiable at $x=-1$?
   Answer: No; a vertical tangent indicates an infinite slope, so $f'(-1)$ does not exist. 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.
6. The graph of $f$ has a corner at $x=0$ and is otherwise smooth; is $f$ continuous at $0$?
   Answer: Yes; a corner can be continuous even though not differentiable. This question tests whether continuity can hold at a corner despite lack of differentiability. A corner at x=0 can still allow the function to be continuous if the left- and right-hand limits match the function value. Continuity depends on limits, not on slopes, so a corner does not force a discontinuity. The smoothness elsewhere supports possible continuity at the corner. Choice A is a tempting distractor, suggesting corners force discontinuities, but it fails because examples like |x| are continuous at corners. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.
7. The graph of $f$ has a cusp at $x=-4$; which statement about $f'(-4)$ is correct?
   Answer: It does not exist because a cusp prevents a well-defined tangent slope. 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.
8. For $f(x)=\begin{cases}x^2,&x<0\\1,&x\ge0\end{cases}$, does $f'(0)$ exist, and why?
   Answer: No; $f$ is discontinuous at $x=0$, so $f'(0)$ does not exist. 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.
9. A function $f$ is continuous at $x=2$ but not differentiable there; which feature could explain this?
   Answer: A corner at $x=2$. This question assesses the connection between differentiability and continuity in AP Calculus AB, identifying features that allow continuity without differentiability. A corner at x=2 maintains continuity but causes unequal one-sided derivatives. This prevents differentiability while preserving continuity. Other options like jumps or holes would break continuity. A tempting distractor might be a hole, but holes cause discontinuity, not fitting the condition. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.
10. A graph of $f$ shows a vertical asymptote at $x=3$; does $f'(3)$ exist, and why?
    Answer: No; $f$ is not defined/continuous at $x=3$, so $f'(3)$ does not exist. 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.
11. The graph of $f$ has a sharp point at $x=8$ and is continuous there; does $f'(8)$ exist?
    Answer: No; a sharp point (corner/cusp) prevents differentiability at $x=8$. 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.
12. For $q(x)=\sqrt[3]{x}$, does $q'(0)$ exist, and why?
    Answer: No; $q$ has a vertical tangent at $0$, so the derivative does not exist. This question assesses differentiability versus continuity, particularly for functions with vertical tangents like the cube root. For q(x) = sqrt[3]{x}, at x=0, the derivative formula gives 1/(3 $x^{2/3}$), which approaches infinity as x approaches 0. This infinite slope means the limit of the difference quotient does not exist finitely, so q'(0) is undefined. The function is continuous at x=0, but differentiability requires a finite derivative. Choice D is a tempting distractor, suggesting cube-root functions have finite slopes everywhere, but it fails because the slope becomes infinite at x=0. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.
13. A graph of $f$ shows a smooth curve except a jump at $x=0$; does $f'(0)$ exist?
    Answer: No; a jump discontinuity means $f$ is not continuous, so $f'(0)$ does not exist. 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.
14. For $f(x)=|x|$, does $f'(0)$ exist, and why?
    Answer: No; there is a corner at $0$, so the one-sided derivatives differ. This question assesses the connection between differentiability and continuity for the absolute value function. For f(x) = |x|, at x=0, the left-hand derivative is -1 and the right-hand is 1, which are unequal. This mismatch means there is no unique derivative at x=0. The function is continuous at x=0, but the corner prevents differentiability. Choice A is a tempting distractor, claiming continuity implies differentiability, but it fails because of the differing slopes. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.
15. A piecewise graph shows $f$ is continuous at $x=2$ but changes slope abruptly; does $f'(2)$ exist?
    Answer: No; an abrupt slope change indicates a corner, so $f'(2)$ does not exist. This question assesses the connection between differentiability and continuity in AP Calculus AB, examining how graph features like corners impact derivatives. The function is continuous at x=2, satisfying the continuity condition. However, an abrupt slope change indicates a corner, where left- and right-hand derivatives differ. This mismatch prevents the two-sided derivative from existing at x=2. A tempting distractor might claim yes due to continuity, but continuity does not guarantee matching one-sided derivatives. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.
16. A function $f$ is differentiable at $x=7$; which statement about continuity at $x=7$ must be true?
    Answer: Yes; differentiability at $x=7$ implies $f$ is continuous at $x=7$. 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.
17. A function $h$ has a cusp at $x=0$; does $h'(0)$ exist, and why?
    Answer: No; a cusp makes slopes unbounded or unequal, so $h'(0)$ does not exist. This question tests understanding of differentiability versus continuity, emphasizing how a cusp affects the derivative. A cusp at x=0 typically means the slopes from the left and right approach infinity or differ in a way that prevents a finite, unique derivative. For h'(0) to exist, the limit of the difference quotient must be finite and the same from both sides, but a cusp violates this by making slopes unbounded or mismatched. The function may still be continuous at the cusp, but that alone does not ensure differentiability. Choice E is a tempting distractor, claiming cusps guarantee differentiability due to continuity, but it fails because cusps often involve infinite or unequal one-sided derivatives. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.
18. For $f(x)=|x-4|+|x-4|^2$, does $f'(4)$ exist, and why?
    Answer: No; the $|x-4|$ term creates a corner at $4$, so $f'(4)$ does not exist. This question assesses the connection between differentiability and continuity in AP Calculus AB, evaluating combined absolute value functions. For f(x) = |x-4| + |x-4|^2, it is continuous at x=4 with value 0. However, the left-hand derivative is -1, while the right-hand is 1, due to the |x-4| term. This mismatch creates a corner, so f'(4) does not exist. A tempting distractor might say yes because of continuity, but the slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.
19. For $g(x)=|x-3|+2$, does $g'(3)$ exist, and why?
    Answer: No; there is a corner at $x=3$, so the derivative does not exist. This question explores the relationship between differentiability and continuity, particularly for absolute value functions that create corners. For g(x) = |x-3| + 2, at x=3, the function changes direction abruptly, forming a corner where the left-hand slope is -1 and the right-hand slope is 1. These unequal one-sided derivatives mean the limit of the difference quotient does not exist, so g'(3) is undefined. Despite the function being continuous at x=3, the mismatch in slopes prevents differentiability. Choice C is a tempting distractor, suggesting continuity implies differentiability, but it fails because continuity is necessary but not sufficient without matching derivatives. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.
20. A graph of $f$ shows a smooth curve crossing $x=1$ but with a hole at $(1,f(1))$; does $f'(1)$ exist?
    Answer: No; a hole means $f$ is discontinuous at $1$, so $f'(1)$ does not exist. 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.

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