Confirming Continuity over an Interval - AP Calculus BC
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Define continuity for a function on a closed interval $[a, b]$.
Define continuity for a function on a closed interval $[a, b]$.
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$f(x)$ is continuous on $[a, b]$ if continuous on $(a, b)$ and limits match at $a$, $b$. Requires continuity at interior points and proper one-sided limits.
$f(x)$ is continuous on $[a, b]$ if continuous on $(a, b)$ and limits match at $a$, $b$. Requires continuity at interior points and proper one-sided limits.
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Is $f(x) = \frac{1}{x-2}$ continuous at $x = 2$?
Is $f(x) = \frac{1}{x-2}$ continuous at $x = 2$?
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No, $f(x)$ is not continuous at $x = 2$. The denominator equals zero, making the function undefined.
No, $f(x)$ is not continuous at $x = 2$. The denominator equals zero, making the function undefined.
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What is $\text{lim}_{x \to 2} (x^2 - 4)/(x - 2)$?
What is $\text{lim}_{x \to 2} (x^2 - 4)/(x - 2)$?
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The limit is $4$, indicating a removable discontinuity. Factor $(x-2)(x+2)$ and cancel to get $\lim_{x \to 2} (x+2) = 4$.
The limit is $4$, indicating a removable discontinuity. Factor $(x-2)(x+2)$ and cancel to get $\lim_{x \to 2} (x+2) = 4$.
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Explain the continuity of $f(x) = \text{sin}(x)$ on $\text{R}$.
Explain the continuity of $f(x) = \text{sin}(x)$ on $\text{R}$.
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$f(x) = \text{sin}(x)$ is continuous on all real numbers. Trigonometric sine function has no domain restrictions.
$f(x) = \text{sin}(x)$ is continuous on all real numbers. Trigonometric sine function has no domain restrictions.
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Is $f(x) = |x|$ continuous at $x = 0$?
Is $f(x) = |x|$ continuous at $x = 0$?
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Yes, $f(x) = |x|$ is continuous at $x = 0$. Both one-sided limits equal 0, matching $f(0) = 0$.
Yes, $f(x) = |x|$ is continuous at $x = 0$. Both one-sided limits equal 0, matching $f(0) = 0$.
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What must be true for a function's one-sided limits at $x = a$ for continuity?
What must be true for a function's one-sided limits at $x = a$ for continuity?
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The one-sided limits must be equal at $x = a$. This ensures the limit exists at the point.
The one-sided limits must be equal at $x = a$. This ensures the limit exists at the point.
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Determine continuity of $f(x) = \frac{1}{x}$ at $x = 0$.
Determine continuity of $f(x) = \frac{1}{x}$ at $x = 0$.
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$f(x)$ is not continuous at $x = 0$ as it is undefined. Division by zero makes the function undefined at this point.
$f(x)$ is not continuous at $x = 0$ as it is undefined. Division by zero makes the function undefined at this point.
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State the three conditions for continuity at a point.
State the three conditions for continuity at a point.
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$f(a)$ is defined, $\text{lim}{x \to a} f(x)$ exists, $\text{lim}{x \to a} f(x) = f(a)$. All three must hold for continuity to be confirmed.
$f(a)$ is defined, $\text{lim}{x \to a} f(x)$ exists, $\text{lim}{x \to a} f(x) = f(a)$. All three must hold for continuity to be confirmed.
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State the limit condition for continuity over an open interval $(a, b)$.
State the limit condition for continuity over an open interval $(a, b)$.
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$f(x)$ is continuous if $\text{lim}_{x \to c} f(x) = f(c)$ for all $c \text{ in } (a, b)$. This is the fundamental definition of continuity over intervals.
$f(x)$ is continuous if $\text{lim}_{x \to c} f(x) = f(c)$ for all $c \text{ in } (a, b)$. This is the fundamental definition of continuity over intervals.
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Determine if $f(x) = 3x^2 - 2x + 1$ is continuous at $x = 2$.
Determine if $f(x) = 3x^2 - 2x + 1$ is continuous at $x = 2$.
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$f(x)$ is continuous at $x = 2$ since it is a polynomial. Polynomials are continuous at every point in their domain.
$f(x)$ is continuous at $x = 2$ since it is a polynomial. Polynomials are continuous at every point in their domain.
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What is the Intermediate Value Theorem?
What is the Intermediate Value Theorem?
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If $f(x)$ is continuous on $[a, b]$, $f(c)$ takes every value between $f(a)$ and $f(b)$. Guarantees existence of intermediate values on continuous functions.
If $f(x)$ is continuous on $[a, b]$, $f(c)$ takes every value between $f(a)$ and $f(b)$. Guarantees existence of intermediate values on continuous functions.
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Determine the continuity of $f(x) = \frac{x}{x^2 - 1}$ at $x = 1$.
Determine the continuity of $f(x) = \frac{x}{x^2 - 1}$ at $x = 1$.
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$f(x)$ is not continuous at $x = 1$ due to division by zero. The denominator $(x-1)(x+1)$ equals zero at $x = 1$.
$f(x)$ is not continuous at $x = 1$ due to division by zero. The denominator $(x-1)(x+1)$ equals zero at $x = 1$.
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Identify the type of discontinuity if $\text{lim}_{x \to a} f(x)$ does not exist.
Identify the type of discontinuity if $\text{lim}_{x \to a} f(x)$ does not exist.
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There is an infinite or jump discontinuity at $x = a$. When one-sided limits differ or approach infinity.
There is an infinite or jump discontinuity at $x = a$. When one-sided limits differ or approach infinity.
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Determine continuity of $f(x) = 2x + 3$ on $(-\text{infinity}, \text{infinity})$.
Determine continuity of $f(x) = 2x + 3$ on $(-\text{infinity}, \text{infinity})$.
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$f(x)$ is continuous for all real numbers. Linear functions are continuous everywhere.
$f(x)$ is continuous for all real numbers. Linear functions are continuous everywhere.
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What is the role of $\text{lim}_{x \to a} f(x)$ in confirming continuity?
What is the role of $\text{lim}_{x \to a} f(x)$ in confirming continuity?
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It must equal $f(a)$ for continuity at $x = a$. The limit must equal the function value for continuity.
It must equal $f(a)$ for continuity at $x = a$. The limit must equal the function value for continuity.
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Is $f(x) = \text{sin}(x)$ continuous at $x = \frac{\text{pi}}{2}$?
Is $f(x) = \text{sin}(x)$ continuous at $x = \frac{\text{pi}}{2}$?
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Yes, $f(x) = \text{sin}(x)$ is continuous at $x = \frac{\text{pi}}{2}$. Sine function is continuous at all points in its domain.
Yes, $f(x) = \text{sin}(x)$ is continuous at $x = \frac{\text{pi}}{2}$. Sine function is continuous at all points in its domain.
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What is the limit condition for $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$?
What is the limit condition for $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$?
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The limit is $2$, indicating a removable discontinuity. Factor and cancel: $\lim_{x \to 1} (x+1) = 2$.
The limit is $2$, indicating a removable discontinuity. Factor and cancel: $\lim_{x \to 1} (x+1) = 2$.
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What is necessary for a function to be continuous on a closed interval $[a, b]$?
What is necessary for a function to be continuous on a closed interval $[a, b]$?
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The function must be continuous on $(a, b)$ and limits must match at $a$ and $b$. Combines interior continuity with proper boundary behavior.
The function must be continuous on $(a, b)$ and limits must match at $a$ and $b$. Combines interior continuity with proper boundary behavior.
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Identify the type of discontinuity in $f(x) = \frac{1}{x^2 - 1}$ at $x = 1$.
Identify the type of discontinuity in $f(x) = \frac{1}{x^2 - 1}$ at $x = 1$.
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Infinite discontinuity at $x = 1$. The function approaches infinity as $x$ approaches 1.
Infinite discontinuity at $x = 1$. The function approaches infinity as $x$ approaches 1.
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What is the continuity status of $f(x) = \text{cos}(x)$?
What is the continuity status of $f(x) = \text{cos}(x)$?
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$f(x) = \text{cos}(x)$ is continuous on all real numbers. Trigonometric cosine function has no domain restrictions.
$f(x) = \text{cos}(x)$ is continuous on all real numbers. Trigonometric cosine function has no domain restrictions.
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Is $f(x) = x^3$ continuous for all real numbers?
Is $f(x) = x^3$ continuous for all real numbers?
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Yes, $f(x) = x^3$ is continuous everywhere. Cubic polynomials have no domain restrictions.
Yes, $f(x) = x^3$ is continuous everywhere. Cubic polynomials have no domain restrictions.
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For $f(x) = \frac{x^2-1}{x-1}$, identify the discontinuity at $x = 1$.
For $f(x) = \frac{x^2-1}{x-1}$, identify the discontinuity at $x = 1$.
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Removable discontinuity at $x = 1$. Factor and cancel to find $\lim_{x \to 1} (x+1) = 2$.
Removable discontinuity at $x = 1$. Factor and cancel to find $\lim_{x \to 1} (x+1) = 2$.
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What is the continuity status of $f(x) = \text{ln}(x)$ at $x = 0$?
What is the continuity status of $f(x) = \text{ln}(x)$ at $x = 0$?
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$f(x) = \text{ln}(x)$ is not continuous at $x = 0$. Natural log is undefined at zero and negative values.
$f(x) = \text{ln}(x)$ is not continuous at $x = 0$. Natural log is undefined at zero and negative values.
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What is the continuity status of $f(x) = \frac{1}{x}$ on $\text{R}$?
What is the continuity status of $f(x) = \frac{1}{x}$ on $\text{R}$?
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$f(x)$ is continuous on $\text{R} \backslash \text{0}$. Continuous everywhere except where the denominator is zero.
$f(x)$ is continuous on $\text{R} \backslash \text{0}$. Continuous everywhere except where the denominator is zero.
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Explain if $f(x) = x^{-2}$ is continuous at $x = 0$.
Explain if $f(x) = x^{-2}$ is continuous at $x = 0$.
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$f(x) = x^{-2}$ is not continuous at $x = 0$. The function is undefined due to division by zero.
$f(x) = x^{-2}$ is not continuous at $x = 0$. The function is undefined due to division by zero.
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What does it mean for $\text{lim}_{x \to a} f(x)$ to exist?
What does it mean for $\text{lim}_{x \to a} f(x)$ to exist?
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Both $\text{lim}{x \to a^-} f(x)$ and $\text{lim}{x \to a^+} f(x)$ exist and are equal. The left and right limits must converge to the same value.
Both $\text{lim}{x \to a^-} f(x)$ and $\text{lim}{x \to a^+} f(x)$ exist and are equal. The left and right limits must converge to the same value.
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Identify the type of discontinuity for a step function.
Identify the type of discontinuity for a step function.
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A step function has a jump discontinuity. Function values change abruptly at certain points.
A step function has a jump discontinuity. Function values change abruptly at certain points.
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Determine the continuity of $f(x) = 5$ on its domain.
Determine the continuity of $f(x) = 5$ on its domain.
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$f(x) = 5$ is continuous everywhere. Constant functions are continuous everywhere.
$f(x) = 5$ is continuous everywhere. Constant functions are continuous everywhere.
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What is the definition of continuity at a point $x = a$?
What is the definition of continuity at a point $x = a$?
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A function $f(x)$ is continuous at $x = a$ if $\text{lim}_{x \to a} f(x) = f(a)$. This ensures the function value equals its limit at that point.
A function $f(x)$ is continuous at $x = a$ if $\text{lim}_{x \to a} f(x) = f(a)$. This ensures the function value equals its limit at that point.
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Can a polynomial function have discontinuities?
Can a polynomial function have discontinuities?
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No, polynomial functions are continuous everywhere. Polynomials have no restrictions on their domain.
No, polynomial functions are continuous everywhere. Polynomials have no restrictions on their domain.
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