Confirming Continuity over an Interval - AP Calculus BC

1. Define $r(x)=\begin{cases}\dfrac{\sin x}{x},&x\ne0\\c,&x=0\end{cases}$; which condition makes $r$ continuous on $[-1,1]$?
   Answer: Set $c=1$ so $\lim_{x\to0}\dfrac{\sin x}{x}=r(0)$ and $r$ is defined on $[-1,1]$. 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.
2. Let $y(x)=\dfrac{\sqrt{x+4}-2}{x}$ for $x\ne0$ and $y(0)=c$; which $c$ makes $y$ continuous on $[-1,1]$?
   Answer: $c=\tfrac{1}{4}$, because rationalizing gives $\lim_{x\to0}y(x)=\tfrac{1}{4}$. 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.
3. For $u(x)=\tan x$ on $\left[-\tfrac{\pi}{3},\tfrac{\pi}{3}\right]$, which statement correctly describes continuity on the interval?
   Answer: $u$ is continuous on the interval because $\cos x\ne0$ for all $x$ in it 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.
4. For $s(x)=|x-1|+\dfrac{1}{x+2}$, which statement about continuity on $[-1,3]$ is correct?
   Answer: $s$ is continuous on $[-1,3]$ because absolute value and rational functions are continuous where defined 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.
5. For $w(x)=\tan x$ restricted to $\left[-\frac{\pi}{3},\frac{\pi}{3}\right]$, which statement verifies continuity on the entire interval?
   Answer: Verify $\cos x\ne0$ on the interval, so $\tan x$ is continuous there. This problem tests confirming continuity over an interval for a trigonometric function with known discontinuities. The function tan(x) = sin(x)/cos(x) is discontinuous where cos(x) = 0, which occurs at x = π/2 + nπ for integer n. On the interval [-π/3, π/3], we need to check if any of these discontinuities occur. Since -π/3 ≈ -1.05 and π/3 ≈ 1.05, while π/2 ≈ 1.57, there are no values where cos(x) = 0 in this interval. Therefore, tan(x) is continuous on the entire interval [-π/3, π/3]. Choice B incorrectly focuses on where sin(x) ≠ 0, which would be relevant for cot(x) = cos(x)/sin(x), not tan(x). To verify trigonometric continuity: identify where the denominator of the ratio form equals zero, check if these points lie in the given interval, and conclude continuity if none are present.
6. Let $m(x)=|x-1|+\dfrac{1}{x-1}$; which statement correctly determines whether $m$ is continuous on $[0,2]$?
   Answer: $m$ is not continuous on $[0,2]$ because it is undefined at $x=1$, an interior point of the interval. This question assesses the skill of confirming a function's continuity over a closed interval. For m(x)=|x-1| + 1/(x-1) to be continuous on [0,2], it must be defined at every point, but at x=1, 1/(x-1) is undefined. The absolute value is continuous everywhere, but the rational term has a vertical asymptote at x=1, an interior point. Limits from left and right at x=1 go to -∞ and +∞, respectively, so no finite limit exists. A tempting distractor is choice C, which claims the discontinuity is removable by simplifying, but no simplification cancels the issue, and it's not removable. 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.
7. Let $p(x)=|x-3|+k$ on $[1,5]$; which conditions confirm $p$ is continuous on $[1,5]$ for any real $k$?
   Answer: Show $|x-3|$ is continuous on $[1,5]$ and adding a constant $k$ preserves continuity; endpoints are defined. 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.
8. A piecewise function is $g(x)=\begin{cases}x+2,&x<1\\ax^2,&x\ge1\end{cases}$; what condition on $a$ makes $g$ continuous on $[-2,3]$?
   Answer: Set $a=3$ so $g(1^-)=g(1)=g(1^+)$ and $g$ is defined on $[-2,3]$. 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.
9. Let $f(x)=\dfrac{x^2-9}{x-3}$ for $x\ne3$ and $f(3)=k$; which condition makes $f$ continuous on $[0,6]$?
   Answer: Choose $k=6$ so $\lim_{x\to3}f(x)=f(3)$ and $f$ is defined on $[0,6]$. This question assesses the skill of confirming a function's continuity over a closed interval. For a function to be continuous on a closed interval like [0,6], it must be defined at every point in the interval, the limit must exist and equal the function value at every interior point, and appropriate one-sided limits must match the function values at the endpoints. In this case, the rational function has a potential discontinuity at x=3, but simplifying shows it equals x+3 for x≠3, so the limit at x=3 is 6. Setting k=6 removes the removable discontinuity, ensuring continuity everywhere on [0,6] since the redefined function is essentially linear. A tempting distractor is choice C, which suggests setting k=0 to cancel (x-3), but that ignores the actual limit value and would create a jump discontinuity. 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.
10. Let $q(x)=\tan(x)$; which statement correctly determines continuity of $q$ on $\left[0,\frac{\pi}{2}\right]$?
    Answer: $q$ is not continuous on $\left[0,\frac{\pi}{2}\right]$ because it is undefined at $x=\frac{\pi}{2}$, so continuity on the closed interval fails. This question assesses the skill of confirming a function's continuity over a closed interval. For q(x)=tan(x) to be continuous on [0, π/2], it must be defined at every point, but tan(π/2) is undefined due to division by zero. While tan(x) is continuous on [0, π/2), the asymptote as x approaches π/2 from the left means no finite limit exists, and it's undefined at the endpoint. Thus, continuity fails on the closed interval. A tempting distractor is choice E, which proposes redefining q(π/2)=0, but the left-hand limit is +∞, not 0, so it wouldn't match. 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.
11. A function $F$ is continuous on $[1,4]$ except possibly at $x=2$, where $\lim_{x\to2}F(x)=5$; what ensures continuity on $[1,4]$?
    Answer: Ensure $F(2)=5$ and $F$ is defined for every $x\in[1,4]$. 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.
12. For $f(x)=\begin{cases}\sqrt{x+2},&-2\le x<1\\x^2-1,&1\le x\le 3\end{cases}$, which conditions confirm $f$ is continuous on $[-2,3]$?
    Answer: Verify $f(-2)$ exists, $f(3)$ exists, and $\lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)=f(1)$. 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 a function to be continuous on a closed interval like [-2,3], it must be defined at every point, the limit must exist at every interior point, and one-sided limits must match the function values at the endpoints. In this piecewise function, the critical point is x=1 where the definition switches, so we must ensure the left-hand limit, right-hand limit, and function value at x=1 are equal. Additionally, verifying that f(-2) and f(3) exist ensures the function is defined at the endpoints, and since each piece is continuous on its subinterval, this confirms overall continuity. A tempting distractor like choice E fails because it only checks that the one-sided limits at x=1 are equal but neglects to verify they equal f(1), which is necessary for continuity at that point. 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.
13. For $h(x)=\sqrt{x-1}+\ln(5-x)$, on which interval is $h$ continuous?
    Answer: $[1,5)$, because $x-1\ge0$ and $5-x>0$. This problem tests confirming continuity over an interval by identifying the domain where a composite function has no breaks. For h(x) = √(x-1) + ln(5-x) to be continuous, both components must be defined and continuous. The square root requires x - 1 ≥ 0, so x ≥ 1. The natural logarithm requires 5 - x > 0, so x < 5. The intersection gives 1 ≤ x < 5, or [1,5). Choice D incorrectly claims ln(5-x) allows x = 5, but ln(0) is undefined. To verify continuity of composite functions on an interval: identify the domain restrictions for each component, find the intersection of valid domains, and check that each component is continuous on the resulting interval.
14. Let $s(x)=\dfrac{1}{x-4}+\dfrac{1}{4-x}$; which statement correctly addresses continuity of $s$ on $[3,5]$?
    Answer: $s$ is not continuous on $[3,5]$ because it is undefined at $x=4$, so it cannot be continuous on the interval. This question assesses the skill of confirming a function's continuity over a closed interval. For s(x)=1/(x-4) + 1/(4-x) to be continuous on [3,5], it must be defined at every point, but at x=4, both terms are undefined. Although simplifying shows s(x)=0 for x≠4 and the limit at x=4 is 0, the function as given isn't defined at this interior point. Therefore, it's not continuous on [3,5] without redefinition. A tempting distractor is choice D, which suggests defining s(4)=0 without checking limits, but while limits match 0, the issue is the original undefined nature at an interior point. 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.
15. If $p$ is polynomial and $r(x)=\dfrac{p(x)}{x^2-4}$, which condition ensures $r$ is continuous on $[-3,3]$?
    Answer: Ensure $p(2)=p(-2)=0$ so the denominator's zeros are removable. 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.
16. Let $r(x)=\sqrt{9-x^2}$ on $[-3,3]$; which conditions confirm $r$ is continuous on $[-3,3]$?
    Answer: Verify $9-x^2\ge0$ on $[-3,3]$ and use continuity of polynomials and $\sqrt{\,}$ on its domain; endpoints are defined. 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 r(x) = √(9 - x²) on [-3,3], verify that 9 - x² ≥ 0 throughout, which holds as it equals zero at endpoints and positive inside. Polynomials are continuous, and the square root is continuous on [0,∞), so their composition is continuous where defined, with endpoints r(-3)=0 and r(3)=0 existing. This ensures no issues in the interval. A tempting distractor like choice E fails because it only verifies the right-hand limit at x=-3 equals r(-3), but continuity requires checks across the entire interval, not just one endpoint. 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.
17. A temperature model is $T(t)=\ln(t+1)$ for $0\le t\le 2$ and $T(t)=at+b$ for $2<t\le 6$; which conditions ensure continuity on $[0,6]$?
    Answer: Choose $a,b$ so that $\lim_{t\to 2^-}T(t)=\lim_{t\to 2^+}T(t)=T(2)$, and ensure $T(0)$ and $T(6)$ exist. 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. To make T continuous on $[0,6]$, we need to choose a and b such that the left-hand limit $\lim_{t \to 2^{-}} T(t)$ equals the right-hand limit $\lim_{t \to 2^{+}} T(t)$ and both equal $T(2)$, ensuring no jump at the piecewise junction. Verifying $T(0)$ and $T(6)$ exist ensures the endpoints are defined, and since $\ln(t+1)$ is continuous on $[0,2]$ and the linear function on $(2,6]$, this covers the interval. The condition $\lim_{t \to 2^{-}} T(t) = \lim_{t \to 2^{+}} T(t) = T(2)$ specifically matches the pieces at t=2. A tempting distractor like choice D fails because it only ensures the one-sided limits at t=2 are equal but ignores equating them to $T(2)$ and does not require endpoints to be defined for closed interval 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.
18. Suppose $q(x)=\frac{1}{x^2-4}$. Which conditions guarantee $q$ is continuous on $[-1,1]$?
    Answer: Confirm $q$ is defined for all $x\in[-1,1]$ and rational functions are continuous on their domains. 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.
19. Let $g(x)=\ln(x-1)+c$ with domain $x>1$; which $c$ makes $g$ continuous on $[2,6]$?
    Answer: Any real $c$, since $\ln(x-1)$ is continuous for $x>1$ 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.
20. For $h(x)=\begin{cases}x^2+ax,&x<1\\3x+b,&x\ge1\end{cases}$, which condition makes $h$ continuous on $[-1,3]$?
    Answer: $1+a=3+b$ 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.

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