Connecting Differentiability and Continuity - AP Calculus BC

1. Suppose $h$ is differentiable at $x=-3$. Which conclusion must follow about $h$ at $x=-3$?
   Answer: $h$ is continuous at $x=-3$. This question directly applies the fundamental theorem connecting differentiability and continuity. If a function is differentiable at a point, then it must be continuous at that point—this is a one-way implication that cannot be reversed. The existence of h'(-3) immediately guarantees that h is continuous at x = -3, making choice A the necessary conclusion. Choice B (corner) would mean the function is not differentiable, contradicting our given information. Choice D incorrectly suggests that continuity depends on the derivative's value being zero, when in fact any finite derivative value ensures continuity. Remember: differentiability is a stronger condition than continuity, so differentiability always implies continuity.
2. If $f$ is continuous at $x=0$ but $f$ has a vertical tangent at $x=0$, which statement is true at $x=0$?
   Answer: $f$ is continuous at $x=0$ but not differentiable at $x=0$. This question explores vertical tangents and their impact on differentiability and continuity. A vertical tangent at x = 0 means the tangent line is vertical (undefined slope), which occurs when $\lim_{x→0}$ |f'(x)| = ∞. This makes f non-differentiable at x = 0 because f'(0) would need to be a finite number. However, the problem states f is continuous at x = 0, which is entirely possible—think of y = ∛x at x = 0, which has a vertical tangent but no break. Choice E incorrectly claims vertical tangents cause discontinuity, but they only prevent differentiability. When you encounter vertical tangents, the function can still be continuous (no break) but will definitely not be differentiable (infinite slope).
3. A function $s$ is continuous at $x=5$ but has a vertical tangent there. Which statement must be true?
   Answer: If $s$ is differentiable at $x=5$, then $s$ is continuous at $x=5$. 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.
4. If $f'(5)$ exists, which statement about $\lim_{x \to 5} f(x)$ and $f(5)$ must be true?
   Answer: $\lim_{x \to 5} f(x) = f(5)$, so $f$ is continuous at $x=5$. This question tests the fundamental theorem that differentiability implies continuity. If $f'(5)$ exists, then $f$ must be continuous at $x = 5$, which means $\lim_{x \to 5} f(x) = f(5)$ must be true. This is because the derivative limit $f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h}$ can only exist if $f(5+h)$ approaches $f(5)$ as $h \to 0$, which is precisely the continuity condition. Differentiability is a stronger requirement than continuity—it demands not just that the function doesn't jump, but that it's smooth enough for a tangent line. Choice B or C would contradict this fundamental relationship. Always remember: if a derivative exists at a point, continuity at that point is guaranteed.
5. If $f(2)=5$ and $\lim_{x\to2}f(x)=5$ but $f'(2)$ does not exist, which statement must be true?
   Answer: $f$ is continuous at $x=2$ but not differentiable at $x=2$. 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.
6. Suppose $q$ is continuous at $x=-2$ but $q'(-2)$ does not exist. Which statement must be true?
   Answer: $q$ is continuous at $x=-2$ but not differentiable at $x=-2$. 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.
7. If $f$ is differentiable on $(1,3)$ and continuous on $[1,3]$, which statement about $f$ at $x=1$ must be true?
   Answer: $f$ must be continuous at $x=1$. 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.
8. For $r(x)=\begin{cases}x^2,&x\ne0\\1,&x=0\end{cases}$, which statement about $r$ at $x=0$ is correct?
   Answer: $r$ is neither continuous nor differentiable at $x=0$. 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 function is not continuous at x=0 due to the limit not equaling the function value. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.
9. Let $u(x)=\begin{cases}\sin x,&x\ne0\\2,&x=0\end{cases}$. Which statement about $u$ at $x=0$ is correct?
   Answer: $u$ is neither continuous nor differentiable at $x=0$. 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.
10. Let $p(x)=|x-1|+2$. Which statement correctly relates continuity and differentiability at $x=1$?
    Answer: $p$ is continuous at $x=1$ but not differentiable at $x=1$. 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.
11. If $f(2)=5$, $\lim_{x\to2}f(x)=5$, and $f'(2)$ does not exist, which statement must be true?
    Answer: $f$ is continuous at $x=2$ but not differentiable at $x=2$. 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 it suggests a function can be differentiable without being continuous, violating the theorem that differentiability implies continuity. To apply this implication strategically, always check continuity first when assessing differentiability, as discontinuity immediately rules out differentiability.
12. Given $t(1)=0$ and $\lim_{x\to1}t(x)$ does not exist, what must be true about $t'(1)$?
    Answer: $t'(1)$ does not exist because differentiability implies continuity. 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.
13. Suppose $r$ is differentiable at $x=a$. Which statement about $r$ at $x=a$ is necessarily true?
    Answer: If $r$ is differentiable at $x=a$, then $r$ is continuous at $x=a$. 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.
14. If $t'(1)$ exists and equals $7$, which statement about $t$ at $x=1$ must be true?
    Answer: $t$ is continuous at $x=1$. 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.
15. Let $q(x)=|x|$. Which statement correctly relates continuity and differentiability of $q$ at $x=0$?
    Answer: $q$ is continuous at $x=0$ but not differentiable at $x=0$. This question uses the absolute value function to illustrate that continuity doesn't guarantee differentiability. The function q(x) = |x| equals -x for x < 0 and x for x ≥ 0. At x = 0, we have q(0) = 0, and both one-sided limits equal 0, so q is continuous at x = 0. However, the left derivative is $\lim_{h→0^-}$ (|h|-0)/h = $\lim_{h→0^-}$ (-h)/h = -1, while the right derivative is $\lim_{h→0^+}$ (|h|-0)/h = $\lim_{h→0^+}$ h/h = 1. Since the one-sided derivatives don't match, q is not differentiable at x = 0 despite being continuous there. Choice D incorrectly claims differentiability without continuity, which violates the fundamental theorem. The absolute value function perfectly demonstrates that a continuous function can fail to be differentiable at points where it has a sharp corner.
16. If $f$ has a corner at $x=4$ but $\lim_{x\to4}f(x)=f(4)$, which statement is correct at $x=4$?
    Answer: $f$ is continuous at $x=4$ but not differentiable at $x=4$. 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.
17. If $f$ has a jump discontinuity at $x=7$, which statement about differentiability and continuity at $x=7$ is true?
    Answer: $f$ is neither continuous at $x=7$ nor differentiable at $x=7$. 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.
18. If $p$ has a jump discontinuity at $x=4$, what must be true about differentiability at $x=4$?
    Answer: $p$ is not differentiable at $x=4$. 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.
19. Given $f(x)=|x-3|$ with $f(3)=0$, which statement best describes $f$ at $x=3$?
    Answer: $f$ is continuous at $x=3$ but not differentiable at $x=3$. 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.
20. Suppose $\lim_{x\to0}f(x)=2$ but $f(0)$ is undefined; which statement about $f$ at $x=0$ is true?
    Answer: $f$ is neither continuous at $x=0$ nor differentiable at $x=0$. This question tests understanding of continuity requirements and their relation to differentiability. For f to be continuous at x = 0, we need three conditions: f(0) must exist, $\lim_{x→0}$ f(x) must exist, and these two must be equal. Here, the limit exists (equals 2) but f(0) is undefined, so f fails the first requirement for continuity at x = 0. Since differentiability requires continuity, and f is not continuous at x = 0, f cannot be differentiable there either. Choice E incorrectly assumes that the existence of a limit implies differentiability, ignoring that the function must first be defined and continuous. When checking differentiability, always verify continuity first—if continuity fails, differentiability automatically fails.

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