Removing Discontinuities - AP Calculus AB
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What is the simplified form of $f(x) = \frac{x^2 - 16}{x - 4}$?
What is the simplified form of $f(x) = \frac{x^2 - 16}{x - 4}$?
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$x + 4$, after factoring and canceling $(x-4)$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.
$x + 4$, after factoring and canceling $(x-4)$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.
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How do you simplify $f(x) = \frac{x^2 - 1}{x - 1}$?
How do you simplify $f(x) = \frac{x^2 - 1}{x - 1}$?
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$x + 1$, after factoring and canceling $(x-1)$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1$.
$x + 1$, after factoring and canceling $(x-1)$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1$.
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Does $f(x) = \frac{x^2 - 4x + 4}{x - 2}$ have a removable discontinuity?
Does $f(x) = \frac{x^2 - 4x + 4}{x - 2}$ have a removable discontinuity?
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Yes, at $x = 2$, as $f(x)$ simplifies to $x - 2$. Factor $(x-2)^2$ in numerator to cancel $(x-2)$ leaving $x-2$.
Yes, at $x = 2$, as $f(x)$ simplifies to $x - 2$. Factor $(x-2)^2$ in numerator to cancel $(x-2)$ leaving $x-2$.
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To remove a discontinuity, what must be true about $\lim_{x \to a} f(x)$?
To remove a discontinuity, what must be true about $\lim_{x \to a} f(x)$?
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It must exist and equal a finite value. Infinite limits cannot be used to remove discontinuities.
It must exist and equal a finite value. Infinite limits cannot be used to remove discontinuities.
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Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x = 3$.
Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x = 3$.
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No, it has a removable discontinuity at $x = 3$. The function is undefined at $x=3$ but has a finite limit there.
No, it has a removable discontinuity at $x = 3$. The function is undefined at $x=3$ but has a finite limit there.
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What does it mean if $\lim_{x \to a} f(x)$ does not exist?
What does it mean if $\lim_{x \to a} f(x)$ does not exist?
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The function has a non-removable discontinuity at $x = a$. Either one-sided limits differ or approach infinity.
The function has a non-removable discontinuity at $x = a$. Either one-sided limits differ or approach infinity.
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What type of discontinuity does a piecewise function often have?
What type of discontinuity does a piecewise function often have?
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Jump discontinuity, if the pieces do not connect. Different function rules at boundaries often create jumps.
Jump discontinuity, if the pieces do not connect. Different function rules at boundaries often create jumps.
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Identify the type of discontinuity in $g(x) = \frac{1}{x - 3}$ at $x = 3$.
Identify the type of discontinuity in $g(x) = \frac{1}{x - 3}$ at $x = 3$.
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Infinite discontinuity at $x = 3$. The denominator approaches zero causing infinite behavior.
Infinite discontinuity at $x = 3$. The denominator approaches zero causing infinite behavior.
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Which discontinuity is not found in $y = x^3 - 3x + 2$?
Which discontinuity is not found in $y = x^3 - 3x + 2$?
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No discontinuities; continuous for all real $x$. Polynomial functions are continuous everywhere with no discontinuities.
No discontinuities; continuous for all real $x$. Polynomial functions are continuous everywhere with no discontinuities.
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Identify the removable discontinuity in $f(x) = \frac{x^2 - 9}{x + 3}$.
Identify the removable discontinuity in $f(x) = \frac{x^2 - 9}{x + 3}$.
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At $x = -3$, as $f(x)$ can be redefined by simplifying. Factor $x^2-9=(x+3)(x-3)$ to cancel $(x+3)$ and get $x-3$.
At $x = -3$, as $f(x)$ can be redefined by simplifying. Factor $x^2-9=(x+3)(x-3)$ to cancel $(x+3)$ and get $x-3$.
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Is $f(x) = x^2 - 4$ continuous for all real numbers?
Is $f(x) = x^2 - 4$ continuous for all real numbers?
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Yes, it is a polynomial with no discontinuities. Polynomial functions have no breaks, holes, or asymptotes.
Yes, it is a polynomial with no discontinuities. Polynomial functions have no breaks, holes, or asymptotes.
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Which discontinuity type cannot be removed by redefining $f(x)$?
Which discontinuity type cannot be removed by redefining $f(x)$?
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Jump and infinite discontinuities. These involve limits that don't exist or are infinite.
Jump and infinite discontinuities. These involve limits that don't exist or are infinite.
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Can all discontinuities be removed?
Can all discontinuities be removed?
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No, only removable discontinuities can be redefined to be continuous. Only holes can be filled; jumps and infinite breaks cannot.
No, only removable discontinuities can be redefined to be continuous. Only holes can be filled; jumps and infinite breaks cannot.
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How can you remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$?
How can you remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$?
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Redefine $f(x)$ at $x = 2$ as $f(x) = x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and get $x+2$.
Redefine $f(x)$ at $x = 2$ as $f(x) = x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and get $x+2$.
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For $f(x) = x^2 - 4$, where is the removable discontinuity?
For $f(x) = x^2 - 4$, where is the removable discontinuity?
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No removable discontinuity; $f(x)$ is continuous. Polynomial functions are continuous everywhere on their domains.
No removable discontinuity; $f(x)$ is continuous. Polynomial functions are continuous everywhere on their domains.
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Identify the type of discontinuity in $f(x) = \frac{1}{x - 2}$ at $x = 2$.
Identify the type of discontinuity in $f(x) = \frac{1}{x - 2}$ at $x = 2$.
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Infinite discontinuity at $x = 2$. The denominator approaches zero causing the function to approach infinity.
Infinite discontinuity at $x = 2$. The denominator approaches zero causing the function to approach infinity.
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What is a removable discontinuity?
What is a removable discontinuity?
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A point where a function is not defined but can be redefined to make it continuous. This describes a hole that can be filled by redefining the function value.
A point where a function is not defined but can be redefined to make it continuous. This describes a hole that can be filled by redefining the function value.
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Which type of discontinuity is present in $f(x) = \frac{1}{x}$ at $x = 0$?
Which type of discontinuity is present in $f(x) = \frac{1}{x}$ at $x = 0$?
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Infinite discontinuity at $x = 0$. The denominator approaches zero while numerator stays constant.
Infinite discontinuity at $x = 0$. The denominator approaches zero while numerator stays constant.
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What is the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$?
What is the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$?
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At $x = 5$, as $f(x)$ can be redefined as $x + 5$. Factor $x^2-25=(x-5)(x+5)$ to cancel $(x-5)$ and get $x+5$.
At $x = 5$, as $f(x)$ can be redefined as $x + 5$. Factor $x^2-25=(x-5)(x+5)$ to cancel $(x-5)$ and get $x+5$.
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What must be true for a function to be continuous at $x = a$?
What must be true for a function to be continuous at $x = a$?
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$f(a) = \lim_{x \to a} f(x)$ and both must exist. All three conditions ensure no gaps, jumps, or holes in the function.
$f(a) = \lim_{x \to a} f(x)$ and both must exist. All three conditions ensure no gaps, jumps, or holes in the function.
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How do you redefine $f(x) = \frac{x^2 - 16}{x - 4}$ to remove the discontinuity?
How do you redefine $f(x) = \frac{x^2 - 16}{x - 4}$ to remove the discontinuity?
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Redefine $f(x)$ at $x = 4$ as $f(x) = x + 4$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.
Redefine $f(x)$ at $x = 4$ as $f(x) = x + 4$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.
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When is a discontinuity non-removable?
When is a discontinuity non-removable?
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If $
exists , \lim_{x \to a} f(x)$; jump or infinite discontinuities. When limits don't exist or are infinite, discontinuities cannot be removed.
If $ exists , \lim_{x \to a} f(x)$; jump or infinite discontinuities. When limits don't exist or are infinite, discontinuities cannot be removed.
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Describe a jump discontinuity.
Describe a jump discontinuity.
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A discontinuity where the left and right limits exist but are unequal. The one-sided limits differ, creating a break in the function.
A discontinuity where the left and right limits exist but are unequal. The one-sided limits differ, creating a break in the function.
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What is a common factor indicating a removable discontinuity?
What is a common factor indicating a removable discontinuity?
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A factor that cancels in both numerator and denominator. Canceling common factors reveals removable discontinuities.
A factor that cancels in both numerator and denominator. Canceling common factors reveals removable discontinuities.
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Find the removable discontinuity in $g(x) = \frac{x^2 - 1}{x - 1}$.
Find the removable discontinuity in $g(x) = \frac{x^2 - 1}{x - 1}$.
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At $x = 1$, as $g(x)$ can be redefined as $x + 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and simplify to $x+1$.
At $x = 1$, as $g(x)$ can be redefined as $x + 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and simplify to $x+1$.
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How can you remove a discontinuity in $f(x) = \frac{x^2 - 9}{x - 3}$?
How can you remove a discontinuity in $f(x) = \frac{x^2 - 9}{x - 3}$?
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Redefine $f(x)$ at $x = 3$ as $f(x) = x + 3$. Factor $x^2-9=(x-3)(x+3)$ to cancel $(x-3)$ and get $x+3$.
Redefine $f(x)$ at $x = 3$ as $f(x) = x + 3$. Factor $x^2-9=(x-3)(x+3)$ to cancel $(x-3)$ and get $x+3$.
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What is the simplified form of $h(x) = \frac{x^2 + x - 6}{x - 2}$?
What is the simplified form of $h(x) = \frac{x^2 + x - 6}{x - 2}$?
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$x + 3$, after factoring numerator and canceling $(x-2)$. Factor $x^2+x-6=(x-2)(x+3)$ to cancel $(x-2)$ and get $x+3$.
$x + 3$, after factoring numerator and canceling $(x-2)$. Factor $x^2+x-6=(x-2)(x+3)$ to cancel $(x-2)$ and get $x+3$.
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Is the function $f(x) = \frac{x^2 - 1}{x - 1}$ continuous at $x = 1$?
Is the function $f(x) = \frac{x^2 - 1}{x - 1}$ continuous at $x = 1$?
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No, it has a removable discontinuity at $x = 1$. The function is undefined at $x=1$ but has a finite limit there.
No, it has a removable discontinuity at $x = 1$. The function is undefined at $x=1$ but has a finite limit there.
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How do you verify a removable discontinuity exists at $x = a$?
How do you verify a removable discontinuity exists at $x = a$?
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Check if $\lim_{x \to a} f(x)$ exists and $f(a)$ is undefined or $\neq \lim_{x \to a} f(x)$. These conditions define exactly when a discontinuity can be removed.
Check if $\lim_{x \to a} f(x)$ exists and $f(a)$ is undefined or $\neq \lim_{x \to a} f(x)$. These conditions define exactly when a discontinuity can be removed.
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How do you remove a discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$?
How do you remove a discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$?
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Redefine $f(x)$ as $f(x) = x + 1$ at $x = 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1=2$.
Redefine $f(x)$ as $f(x) = x + 1$ at $x = 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1=2$.
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