AP Calculus BC Quiz: Confirming Continuity Over An Interval

What this quiz covers

This quiz focuses on Confirming Continuity Over An Interval, 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. Ensure $p(0)=0$ so the numerator vanishes at the midpoint.
2. Ensure $p(2)=p(-2)=0$ so the denominator’s zeros are removable. (correct answer)
3. Ensure $p$ is differentiable on $[-3,3]$.
4. Ensure $p(x)$ has degree at least $2$.
5. No additional condition; all rational functions are continuous on closed intervals.

Explanation: This problem tests confirming continuity over an interval for a rational function with potential discontinuities. The function r(x) = p(x)/(x² - 4) has discontinuities where the denominator equals zero: x² - 4 = 0 gives x = ±2. For r to be continuous on [-3,3], which includes both x = -2 and x = 2, these must be removable discontinuities. This requires p(x) to have (x - 2) and (x + 2) as factors, meaning p(2) = 0 and p(-2) = 0. Choice A incorrectly focuses only on x = 0, missing the actual discontinuity locations. To ensure rational function continuity on an interval: find where the denominator is zero, verify these points are in the interval, and require the numerator to vanish at these points to make discontinuities removable.

Question 2

1. Show $|x-3|$ is continuous on $[1,5]$ and adding a constant $k$ preserves continuity; endpoints are defined. (correct answer)
2. Show $p$ is differentiable at $x=3$ for all $k$ and continuous elsewhere.
3. Show $p(1)=p(5)$ and $\lim_{x\to 3}p(x)=0$.
4. Show $\lim_{x\to 1}p(x)$ and $\lim_{x\to 5}p(x)$ exist; interior continuity is not required.
5. Show $p$ is continuous only at $x=3$, since absolute value can be discontinuous elsewhere.

Explanation: Confirming continuity over an interval is a key skill in AP Calculus BC that requires checking the function's behavior at every point in the interval, including endpoints and potential discontinuities. The absolute value function |x-3| is continuous everywhere, including on [1,5], as it is composed of linear pieces that meet at x=3. Adding a constant k preserves continuity since constants are continuous, and verifying endpoints are defined ensures p(1) and p(5) exist. This confirms p is continuous for any real k on the interval. A tempting distractor like choice E fails because it suggests continuity only at x=3 is needed, but continuity on the entire interval requires it at every point, not just the vertex. To check continuity on any closed interval [a,b], ensure the function is defined everywhere, limits equal function values at interior points, and appropriate one-sided limits match at endpoints.

Question 3

1. $s$ is continuous on $[-1,3]$ because absolute value and rational functions are continuous where defined (correct answer)
2. $s$ is not continuous on $[-1,3]$ because $|x-1|$ has a cusp at $x=1$
3. $s$ is not continuous on $[-1,3]$ because $\dfrac{1}{x+2}$ is undefined at $x=-2$
4. $s$ is continuous on $[-1,3]$ only if $s'(1)$ exists
5. $s$ is not continuous on $[-1,3]$ because $\lim_{x\to-1}s(x)$ does not exist

Explanation: This problem tests understanding of continuity for functions involving absolute values and rational expressions. For s(x) = |x-1| + 1/(x+2) to be continuous on [-1,3], we need to check potential discontinuities. The absolute value function |x-1| is continuous everywhere (including at x=1 where it has a corner but not a discontinuity), and 1/(x+2) is continuous wherever x≠-2. Since x=-2 is outside the interval [-1,3], the rational part has no discontinuities on our interval. Therefore, s is continuous on [-1,3]. Choice B incorrectly confuses non-differentiability with discontinuity; corners are continuous but not smooth. For checking continuity of combined functions: identify discontinuities of each component, verify these points relative to the interval, and remember that absolute values create corners, not jumps.

Question 4

1. Confirm $q$ is defined for all $x\in[-1,1]$ and rational functions are continuous on their domains. (correct answer)
2. Confirm $q$ is defined for all real $x$ and has no vertical asymptotes.
3. Confirm $\lim_{x\to-1}q(x)$ and $\lim_{x\to1}q(x)$ both equal $0$.
4. Confirm $q(-1)=q(1)$ and $q$ is increasing on $[-1,1]$.
5. Confirm $q$ is continuous at $x=0$ only.

Explanation: This question tests confirming continuity over an interval for a rational function. The function q(x) = 1/(x²-4) has vertical asymptotes where x²-4 = 0, which gives x = ±2. Since neither x = 2 nor x = -2 lies in the interval [-1,1], the function is defined throughout [-1,1]. Rational functions are continuous wherever they're defined, so q is continuous on [-1,1]. Choice C incorrectly focuses on endpoint limits equaling 0, which isn't necessary for continuity. To verify continuity of rational functions on an interval: (1) find where the denominator equals zero (vertical asymptotes), (2) check if any asymptotes lie within your interval, (3) if the interval avoids all asymptotes, the function is continuous there.

Question 5

1. Set $c=0$ so $r(0)=0$ and $\sin 0=0$.
2. Set $c=1$ so $\lim_{x\to0}\dfrac{\sin x}{x}=r(0)$ and $r$ is defined on $[-1,1]$. (correct answer)
3. Any $c$ works because $\dfrac{\sin x}{x}$ is continuous for $x\ne0$.
4. No $c$ works because $\dfrac{\sin x}{x}$ has an infinite discontinuity at $0$.
5. Set $c=\sin(1)$ so endpoint behavior matches at $x=1$.

Explanation: This question assesses the skill of confirming a function's continuity over a closed interval. For r(x) to be continuous on [-1,1], it must be defined everywhere, with the limit at x=0 equaling c, where lim_{x→0} (sin x)/x =1. Setting c=1 fills the removable discontinuity at x=0, and since sin x / x is continuous elsewhere, the whole function is continuous on [-1,1]. The function is well-behaved at the endpoints r(-1)=sin(-1)/(-1) and r(1)=sin(1)/1. A tempting distractor is choice A, which sets c=0 because sin(0)=0, but it ignores the actual limit value of 1. To check continuity on any closed interval [a,b], verify the function is defined throughout, limits equal function values at interiors, and one-sided limits match at endpoints.

Question 6

1. Require $s(2)$ is defined; then $s$ is continuous on $[0,5]$.
2. Require $\lim_{x\to2}s(x)$ exists; then $s$ is continuous on $[0,5]$.
3. Require $\lim_{x\to2^-}s(x)=\lim_{x\to2^+}s(x)=s(2)$. (correct answer)
4. Require $s'(2)$ exists; then $s$ is continuous on $[0,5]$.
5. Require $s(0)=s(5)$ to avoid an endpoint discontinuity.

Explanation: This problem tests confirming continuity over an interval by identifying the complete continuity condition at a point. If s is continuous everywhere on [0,5] except possibly at x = 2, then to guarantee continuity on the entire interval, we need continuity at x = 2. This requires three conditions: s(2) is defined, lim(x→2) s(x) exists, and lim(x→2) s(x) = s(2). Choice C correctly states this as lim(x→2⁻) s(x) = lim(x→2⁺) s(x) = s(2). Choice A only requires s(2) to be defined, missing the limit requirement, while Choice B only requires the limit to exist, missing the equality with s(2). To confirm point continuity: verify the function is defined at the point, check that left and right limits exist and are equal, and ensure this common limit equals the function value.

Question 7

1. On $(-\infty,2)\cup(2,\infty)$, since only $\frac{1}{x-2}$ is discontinuous at $2$. (correct answer)
2. On $(-\infty,\infty)$, since integrals always produce continuous functions.
3. On $[1,\infty)$ only, since the integral starts at $1$.
4. On $(-\infty,2]\cup(2,\infty)$, since continuity can include $x=2$ by definition.
5. Only on $(2,\infty)$, since the integrand is undefined for $t<0$.

Explanation: This problem tests confirming continuity over an interval for a function combining an integral and a rational expression. The integral ∫₁ˣ 1/(t² + 1) dt is continuous for all x because the integrand is continuous everywhere. However, the term 1/(x - 2) has a discontinuity at x = 2 where the denominator is zero. Therefore, F(x) is continuous on (-∞, 2) ∪ (2, ∞). Choice B incorrectly claims integrals always produce continuous functions, ignoring the additional term with a discontinuity. To verify continuity of combined functions: check each component separately, identify all discontinuities, and express the continuous interval as the complement of discontinuity points.

Question 8

1. $(-\infty,\infty)$, since absolute value is continuous.
2. $(-\infty,-1)\cup(-1,\infty)$, excluding where the denominator is zero. (correct answer)
3. $(-\infty,4)\cup(4,\infty)$, excluding the corner at $x=4$.
4. $[-1,\infty)$, because continuity fails only to the left of $-1$.
5. $(-\infty,-1]$, because continuity can include $x=-1$.

Explanation: This problem tests confirming continuity over an interval for a sum of functions with different types of potential discontinuities. The function m(x) = |x - 4| + 1/(x + 1) combines an absolute value (continuous everywhere) with a rational function. The absolute value |x - 4| is continuous for all x, including at the corner point x = 4. The term 1/(x + 1) has a discontinuity only where x + 1 = 0, which is x = -1. Therefore, m is continuous on (-∞, -1) ∪ (-1, ∞). Choice C incorrectly identifies x = 4 as a discontinuity, confusing non-differentiability with discontinuity. To confirm continuity of sums: check each term independently, remember that absolute values are continuous everywhere (though not differentiable at corners), and combine the continuous regions.

Question 9

1. $c=0$, because the numerator is $0$ when $x=0$.
2. $c=\tfrac{1}{4}$, because rationalizing gives $\lim_{x\to0}y(x)=\tfrac{1}{4}$. (correct answer)
3. $c=\tfrac{1}{2}$, because $\sqrt{x+4}$ has derivative $\tfrac{1}{2}$ at $x=0$.
4. $c=2$, because continuity requires matching the square root value at $0$.
5. Any real $c$, because $y$ is continuous for all $x\ne0$.

Explanation: This problem tests confirming continuity over an interval by evaluating a limit using algebraic techniques. For y(x) = (√(x+4) - 2)/x with x ≠ 0, we need to find lim(x→0) y(x) to determine c. Multiply by the conjugate: y(x) = [(√(x+4) - 2)/x] · [(√(x+4) + 2)/(√(x+4) + 2)] = [(x+4) - 4]/[x(√(x+4) + 2)] = x/[x(√(x+4) + 2)] = 1/(√(x+4) + 2). As x → 0: lim(x→0) y(x) = 1/(√4 + 2) = 1/4. Therefore, c = 1/4 makes y continuous at x = 0. Choice C incorrectly uses the derivative of √(x+4) at x = 0, confusing continuity with differentiability. To handle limits with radicals: rationalize by multiplying by the conjugate, simplify the resulting expression, and evaluate the limit of the simplified form.

Question 10

1. $h$ is continuous on $[2,5]$ because logarithms are continuous for all real inputs.
2. $h$ is not continuous on $[2,5]$ because it is undefined at $x=2$, so it cannot be continuous on the interval. (correct answer)
3. $h$ is continuous on $[2,5]$ because it is continuous on $(2,5]$ and endpoints do not matter.
4. $h$ is not continuous on $[2,5]$ because $\ln$ has a removable discontinuity at $x=2$.
5. $h$ is continuous on $[2,5]$ if $\lim_{x\to2^+}h(x)=0$.

Explanation: This question assesses the skill of confirming a function's continuity over a closed interval. For h(x)=ln(x-2) to be continuous on [2,5], it must be defined at every point, including x=2, where ln(0) is undefined. Even though the logarithm is continuous for x>2, the lack of definition at the endpoint x=2 prevents continuity on the closed interval. One-sided limits at endpoints are required, but since h(2) doesn't exist, continuity fails. A tempting distractor is choice E, which suggests checking the right-hand limit at x=2 equals 0, but without defining h(2), the function isn't continuous on [2,5]. To check continuity on any closed interval [a,b], verify the function is defined throughout, limits equal function values at interiors, and one-sided limits match at endpoints.

Question 11

1. Ensure $F(2)=5$ and $F$ is defined for every $x\in[1,4]$. (correct answer)
2. Ensure $F(2)\ne5$ so the limit is not equal to the function value.
3. Ensure $F$ is differentiable at $x=2$; continuity then follows automatically.
4. Ensure only that $F(1)$ and $F(4)$ are defined; interior points need not be checked.
5. Ensure $\lim_{x\to2^-}F(x)=5$; the right-hand limit is unnecessary.

Explanation: This question assesses the skill of confirming a function's continuity over a closed interval. For F to be continuous on [1,4], it must be defined everywhere in the interval, and at x=2, where the limit is 5, setting F(2)=5 ensures the limit equals the function value. Assuming F is continuous elsewhere and defined throughout, this resolves the potential issue at x=2. Continuity requires checking all points, but the given limit condition focuses on x=2. A tempting distractor is choice D, which says only endpoints need definition, but interior points like x=2 must also satisfy continuity conditions. To check continuity on any closed interval [a,b], verify the function is defined throughout, limits equal function values at interiors, and one-sided limits match at endpoints.

Question 12

1. Any real $c$, since $\ln(x-1)$ is continuous for $x>1$ (correct answer)
2. $c$ must satisfy $g(2)=0$
3. $c$ must satisfy $g(6)=0$
4. $c$ must be positive
5. $c$ must be $0$ because endpoints require $c=0$

Explanation: This problem asks about confirming continuity of g(x) = ln(x - 1) + c over the interval [2, 6]. The natural logarithm function ln(x - 1) is continuous for all x > 1, which includes the entire interval [2, 6]. Adding a constant c to a continuous function preserves continuity, so g(x) is continuous on [2, 6] for any real value of c. Choice B (c must satisfy g(2) = 0) incorrectly assumes we need to force a specific value at an endpoint, but continuity doesn't require any particular function values. To verify interval continuity: ensure the function's domain includes the entire interval, check that all component functions are continuous on that interval, and remember that adding constants doesn't affect continuity.

Question 13

1. $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
2. $\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$ (correct answer)
3. $\left[0,\pi\right]$
4. $\left[\dfrac{\pi}{3},\dfrac{2\pi}{3}\right]$
5. $\left[-\pi,\pi\right]$

Explanation: This problem asks for an interval where t(x) = tan(x) is continuous everywhere. The tangent function has vertical asymptotes (discontinuities) at x = π/2 + nπ for any integer n, because tan(x) = sin(x)/cos(x) and cos(x) = 0 at these points. We need an interval that avoids all such discontinuities. Choice A [-π/2, π/2] includes the discontinuity at x = π/2, so tan(x) is not continuous there. Choice B [-π/4, π/4] avoids all discontinuities since the nearest ones are at ±π/2, making tan(x) continuous throughout this interval. To verify continuity of trigonometric functions on intervals: identify where denominators equal zero or where the function is undefined, ensure the interval avoids all such points, and remember that tan(x) has period π with discontinuities at odd multiples of π/2.

Question 14

1. Continuous on $[0,5]$ for all real $m$ (correct answer)
2. Continuous on $[0,5]$ only if $m\ge0$
3. Discontinuous at $x=2$ unless $m=0$
4. Discontinuous at $x=0$ unless $m=2$
5. Discontinuous at $x=5$ unless $m=3$

Explanation: This problem involves confirming continuity of r(x) = |x - 2| + m on the interval [0, 5]. The absolute value function |x - 2| is continuous everywhere, including at x = 2 where it has a corner (the function is continuous but not differentiable there). Adding a constant m to a continuous function preserves continuity, so r(x) is continuous on [0, 5] for any real value of m. Choice C (discontinuous at x = 2 unless m = 0) confuses non-differentiability with discontinuity—absolute value functions have corners but remain continuous. To verify continuity of functions with absolute values: remember that |f(x)| is continuous wherever f(x) is continuous, corners don't create discontinuities, and adding constants preserves continuity.

Question 15

1. $a+b=0$
2. $1+a=3+b$ (correct answer)
3. $a=3$
4. $b=0$
5. $a=b$

Explanation: This problem involves confirming continuity of a piecewise function over the interval [-1, 3]. The function h(x) has two pieces that meet at x = 1, which is inside our interval. For h to be continuous on [-1, 3], it must be continuous at x = 1, meaning the left and right limits must equal: lim(x→1⁻) h(x) = lim(x→1⁺) h(x). From the left: lim(x→1⁻) (x² + ax) = 1 + a, and from the right: lim(x→1⁺) (3x + b) = 3 + b. Setting these equal gives 1 + a = 3 + b. Choice A (a + b = 0) is incorrect because it doesn't ensure the limits match at x = 1. For piecewise continuity: identify transition points within the interval, compute left and right limits at each transition, and set them equal to find the required relationship between parameters.

Question 16

1. Set $a=3$ so $g(1^-)=g(1)=g(1^+)$ and $g$ is defined on $[-2,3]$. (correct answer)
2. Set $a=1$ so both pieces are polynomials and therefore continuous everywhere.
3. Any $a$ works because each piece is continuous on its own subinterval.
4. Set $a=0$ so $g(1)=0$ and the one-sided limits at $1$ must differ.
5. No value of $a$ works because continuity on $[-2,3]$ requires differentiability at $x=1$.

Explanation: This question assesses the skill of confirming a function's continuity over a closed interval. For a piecewise function to be continuous on [-2,3], it must be defined everywhere, with limits matching function values at all points, including the join at x=1 where left and right limits must equal g(1). The left limit at x=1 is $1+2=3$, and the right limit is $a(1)^2=a$, so setting $a=3$ ensures continuity at x=1. Both pieces are polynomials, hence continuous on their domains, and with $a=3$, the whole function is continuous on [-2,3]. A tempting distractor is choice B, which sets $a=1$ thinking polynomials are always continuous, but it fails because the limits at x=1 don't match $(3≠1)$. To check continuity on any closed interval [a,b], verify the function is defined throughout, limits equal function values at interiors, and one-sided limits match at endpoints.

Question 17

1. Verify $\lim_{x\to 0}h(x)=h(0)$ and that $h$ is continuous on $[-1,0)$ and $(0,1]$, with $h(-1)$ and $h(1)$ defined. (correct answer)
2. Verify $h'(0)$ exists and $h$ is bounded on $[-1,1]$.
3. Verify $\lim_{x\to 0}h(x)=0$ and $h(0)=0$.
4. Verify $\lim_{x\to -1}h(x)=h(-1)$ only, since interior points are automatically continuous.
5. Verify $\lim_{x\to 0^-}h(x)=\lim_{x\to 0^+}h(x)$ only.

Explanation: Confirming continuity over an interval is a key skill in AP Calculus BC that requires checking the function's behavior at every point in the interval, including endpoints and potential discontinuities. For h to be continuous on [-1,1], we must verify that lim x→0 h(x) = h(0), which it does since the limit is 1, and ensure h is continuous on the subintervals [-1,0) and (0,1] where sin x / x is naturally continuous. Verifying h(-1) and h(1) are defined addresses the endpoints, with one-sided limits matching from within. This comprehensive check confirms no breaks in the interval. A tempting distractor like choice E fails because it only checks that left and right limits at x=0 are equal but does not verify they equal h(0), which is essential for continuity. To check continuity on any closed interval [a,b], ensure the function is defined everywhere, limits equal function values at interior points, and appropriate one-sided limits match at endpoints.

Question 18

1. $a=0$, since $g(3)$ must equal the numerator at $x=3$.
2. $a=3$, since the factor $x-3$ cancels leaving $x+3$.
3. $a=6$, since $\lim_{x\to3}g(x)=6$. (correct answer)
4. $a=9$, since $\lim_{x\to3}g(x)=9$.
5. Any real $a$, since redefining one point never affects continuity.

Explanation: This problem tests confirming continuity over an interval by finding the value that removes a removable discontinuity. The function g(x) = (x² - 9)/(x - 3) has a removable discontinuity at x = 3 because both numerator and denominator are zero there. Factoring: g(x) = (x - 3)(x + 3)/(x - 3) = x + 3 for x ≠ 3. Therefore, lim(x→3) g(x) = 3 + 3 = 6. For continuity at x = 3, we need g(3) = a = 6. Choice A incorrectly suggests a = 0 by evaluating only the numerator at x = 3, missing that we need the limit of the simplified expression. To confirm continuity on an interval containing a removable discontinuity: simplify the function by canceling common factors, evaluate the limit at the problematic point, and set the function value equal to this limit.

Question 19

1. $k=0$
2. $k=3$
3. $k=6$ (correct answer)
4. $k=9$
5. $k$ can be any real number

Explanation: This problem tests confirming continuity over an interval by finding the value that removes a removable discontinuity. The function f(x) = (x² - 9)/(x - 3) has a discontinuity at x = 3, which lies within the interval [-2, 5]. For x ≠ 3, we can factor: f(x) = (x - 3)(x + 3)/(x - 3) = x + 3. To make f continuous at x = 3, we need f(3) = lim(x→3) f(x) = lim(x→3) (x + 3) = 6. Choice A (k = 0) is tempting if you incorrectly think setting f(3) = 0 somehow "cancels" the discontinuity, but continuity requires matching the limit value. To confirm continuity over an interval: check the function is continuous at all interior points, verify any removable discontinuities are properly filled, and ensure the function is defined at the endpoints.

Question 20

1. $u$ is continuous on the interval because $\cos x\ne0$ for all $x$ in it (correct answer)
2. $u$ is not continuous because $\tan x$ is never continuous at $x=0$
3. $u$ is not continuous because the interval contains $x=\tfrac{\pi}{2}$
4. $u$ is continuous only if $u\left(\tfrac{\pi}{3}\right)=u\left(-\tfrac{\pi}{3}\right)$
5. $u$ is continuous only if $u'(x)$ exists at every point

Explanation: This problem tests understanding of trigonometric function continuity on restricted domains. For u(x) = tan(x) = sin(x)/cos(x) to be continuous on [-π/3, π/3], we need cos(x) ≠ 0 throughout the interval. The tangent function has vertical asymptotes where cos(x) = 0, which occurs at x = π/2 + nπ. The nearest asymptote to our interval is at x = π/2 ≈ 1.57, which is outside [-π/3, π/3] ≈ [-1.05, 1.05]. Therefore, tan(x) is continuous on the entire interval. Choice C incorrectly claims the interval contains π/2, but π/2 > π/3. For trigonometric continuity: identify where denominators equal zero, check if these points lie in the given interval, and remember that tan(x) is continuous wherever cos(x) ≠ 0.