Confirming Continuity over an Interval
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AP Calculus BC › Confirming Continuity over an Interval
Define $r(x)=\begin{cases}\dfrac{\sin x}{x},&x\ne0\\c,&x=0\end{cases}$; which condition makes $r$ continuous on $-1,1$?
Any $c$ works because $\dfrac{\sin x}{x}$ is continuous for $x\ne0$.
No $c$ works because $\dfrac{\sin x}{x}$ has an infinite discontinuity at $0$.
Set $c=0$ so $r(0)=0$ and $\sin 0=0$.
Set $c=\sin(1)$ so endpoint behavior matches at $x=1$.
Set $c=1$ so $\lim_{x\to0}\dfrac{\sin x}{x}=r(0)$ and $r$ is defined on $[-1,1]$.
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.
For $F(x)=\int_{1}^{x}\dfrac{1}{t^2+1},dt+\dfrac{1}{x-2}$, where is $F$ continuous?
On $(-\infty,2]\cup(2,\infty)$, since continuity can include $x=2$ by definition.
On $(-\infty,\infty)$, since integrals always produce continuous functions.
On $[1,\infty)$ only, since the integral starts at $1$.
On $(-\infty,2)\cup(2,\infty)$, since only $\frac{1}{x-2}$ is discontinuous at $2$.
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.
Let $m(x)=|x-4|+\dfrac{1}{x+1}$; on which interval is $m$ continuous?
$[-1,\infty)$, because continuity fails only to the left of $-1$.
$(-\infty,-1)\cup(-1,\infty)$, excluding where the denominator is zero.
$(-\infty,\infty)$, since absolute value is continuous.
$(-\infty,-1]$, because continuity can include $x=-1$.
$(-\infty,4)\cup(4,\infty)$, excluding the corner at $x=4$.
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.
If $p$ is polynomial and $r(x)=\dfrac{p(x)}{x^2-4}$, which condition ensures $r$ is continuous on $-3,3$?
Ensure $p$ is differentiable on $[-3,3]$.
No additional condition; all rational functions are continuous on closed intervals.
Ensure $p(2)=p(-2)=0$ so the denominator’s zeros are removable.
Ensure $p(x)$ has degree at least $2$.
Ensure $p(0)=0$ so the numerator vanishes at the midpoint.
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.
Suppose $s$ is continuous on $0,5$ except possibly at $x=2$; which condition guarantees continuity on $0,5$?
Require $s'(2)$ exists; then $s$ is continuous on $[0,5]$.
Require $s(2)$ is defined; then $s$ is continuous on $[0,5]$.
Require $\lim_{x\to2^-}s(x)=\lim_{x\to2^+}s(x)=s(2)$.
Require $\lim_{x\to2}s(x)$ exists; then $s$ is continuous on $[0,5]$.
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.
Let $y(x)=\dfrac{\sqrt{x+4}-2}{x}$ for $x\ne0$ and $y(0)=c$; which $c$ makes $y$ continuous on $-1,1$?
$c=0$, because the numerator is $0$ when $x=0$.
$c=2$, because continuity requires matching the square root value at $0$.
$c=\tfrac{1}{4}$, because rationalizing gives $\lim_{x\to0}y(x)=\tfrac{1}{4}$.
Any real $c$, because $y$ is continuous for all $x\ne0$.
$c=\tfrac{1}{2}$, because $\sqrt{x+4}$ has derivative $\tfrac{1}{2}$ at $x=0$.
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.
Let $h(x)=\ln(x-2)$; which statement correctly justifies whether $h$ is continuous on $2,5$?
$h$ is continuous on $[2,5]$ because it is continuous on $(2,5]$ and endpoints do not matter.
$h$ is not continuous on $[2,5]$ because it is undefined at $x=2$, so it cannot be continuous on the interval.
$h$ is continuous on $[2,5]$ because logarithms are continuous for all real inputs.
$h$ is continuous on $[2,5]$ if $\lim_{x\to2^+}h(x)=0$.
$h$ is not continuous on $[2,5]$ because $\ln$ has a removable discontinuity at $x=2$.
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.
Let $p(x)=|x-3|+k$ on $1,5$; which conditions confirm $p$ is continuous on $1,5$ for any real $k$?
Show $p$ is differentiable at $x=3$ for all $k$ and continuous elsewhere.
Show $|x-3|$ is continuous on $[1,5]$ and adding a constant $k$ preserves continuity; endpoints are defined.
Show $p$ is continuous only at $x=3$, since absolute value can be discontinuous elsewhere.
Show $\lim_{x\to 1}p(x)$ and $\lim_{x\to 5}p(x)$ exist; interior continuity is not required.
Show $p(1)=p(5)$ and $\lim_{x\to 3}p(x)=0$.
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.
A function $F$ is continuous on $1,4$ except possibly at $x=2$, where $\lim_{x\to2}F(x)=5$; what ensures continuity on $1,4$?
Ensure $F(2)=5$ and $F$ is defined for every $x\in[1,4]$.
Ensure $F(2)\ne5$ so the limit is not equal to the function value.
Ensure only that $F(1)$ and $F(4)$ are defined; interior points need not be checked.
Ensure $F$ is differentiable at $x=2$; continuity then follows automatically.
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.
Let $g(x)=\ln(x-1)+c$ with domain $x>1$; which $c$ makes $g$ continuous on $2,6$?
$c$ must satisfy $g(6)=0$
$c$ must be $0$ because endpoints require $c=0$
Any real $c$, since $\ln(x-1)$ is continuous for $x>1$
$c$ must be positive
$c$ must satisfy $g(2)=0$
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.