What is a removable discontinuity?

A point where the limit exists, but the function is undefined or has a different value.

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What is a removable discontinuity?

A point where the limit exists, but the function is undefined or has a different value.

Define continuity at a point.

A function is continuous at x=a if the limit as x approaches a exists, f(a) is defined, and the limit equals f(a).

What is a piecewise function?

A function defined by multiple sub-functions, each applying to a certain interval of the domain.

Define a limit of a function.

The value that a function approaches as the input approaches some value.

What does it mean for a function to be undefined at a point?

The function does not have a value at that specific x-value.

What is the domain of a function?

The set of all possible input values (x-values) for which the function is defined.

What is a continuous function?

A function that can be drawn without lifting your pencil.

Define left-hand limit.

The value the function approaches as x approaches a value from the left.

Define right-hand limit.

The value the function approaches as x approaches a value from the right.

What is meant by 'filling the gap' in the context of removable discontinuities?

Redefining the function's value at the point of discontinuity to be equal to the limit at that point, thus making the function continuous.

How to find and remove a discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$ ?

Factor the numerator: $f(x) = \frac{(x - 2)(x + 2)}{x - 2}$ . Cancel common factors: $f(x) = x + 2$ . Redefine $f(2) = 4$ .

How to ensure continuity of $f(x) = \begin{cases} x^2, & x \leq 1 \\ ax, & x > 1 \end{cases}$ ?

Set $x^2 = ax$ at $x = 1$ . Solve for $a$ : $1^2 = a(1)$ , so $a = 1$ .

How to determine if $f(x) = \frac{x^2 - 1}{x + 1}$ has a removable discontinuity?

Factor: $f(x) = \frac{(x - 1)(x + 1)}{x + 1}$ . Cancel: $f(x) = x - 1$ . Yes, at $x = -1$ .

How to redefine $f(x) = \frac{x^2 - 9}{x - 3}$ to be continuous?

Factor: $f(x) = \frac{(x - 3)(x + 3)}{x - 3}$ . Cancel: $f(x) = x + 3$ . Define $f(3) = 6$ .

How to find the value of 'a' to make $f(x) = \begin{cases} x + a, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ continuous?

Set $x + a = x^2$ at $x = 0$ . Solve for $a$ : $0 + a = 0^2$ , so $a = 0$ .

Given $f(x) = \frac{x^2 - 4x + 3}{x - 1}$ , how do you find and remove the discontinuity?

Factor the numerator: $f(x) = \frac{(x - 1)(x - 3)}{x - 1}$ . Cancel the common factor: $f(x) = x - 3$ . Evaluate at x = 1: $f(1) = 1 - 3 = -2$ .

How do you ensure the function $f(x) = \begin{cases} cx^2, & x \leq 2 \\ 2x + c, & x > 2 \end{cases}$ is continuous?

Set the two pieces equal at x = 2: $c(2)^2 = 2(2) + c$ . Solve for c: $4c = 4 + c$ , so $3c = 4$ , and $c = \frac{4}{3}$ .

How do you find the limit of $f(x) = \frac{x^2 - 25}{x + 5}$ as x approaches -5?

Factor the numerator: $f(x) = \frac{(x - 5)(x + 5)}{x + 5}$ . Cancel the common factor: $f(x) = x - 5$ . Evaluate at x = -5: $f(-5) = -5 - 5 = -10$ .

How do you find the value of 'k' that makes $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases}$ continuous?

Factor the numerator: $f(x) = \frac{(x - 3)(x + 3)}{x - 3}$ . Cancel the common factor: $f(x) = x + 3$ . Evaluate at x = 3: $f(3) = 3 + 3 = 6$ . Set k = 6.

How do you determine if the function $f(x) = \frac{x^2 + 3x + 2}{x + 2}$ has a removable discontinuity and remove it?

Factor the numerator: $f(x) = \frac{(x + 1)(x + 2)}{x + 2}$ . Cancel the common factor: $f(x) = x + 1$ . Yes, it has a removable discontinuity at x = -2. Evaluate at x = -2: $f(-2) = -2 + 1 = -1$ .

How do you test continuity for a piecewise function at x=a?

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

What is the general form of a rational function where removable discontinuities often occur?

$f(x) = \frac{p(x)}{q(x)}$ , where p(x) and q(x) are polynomials.

How do you find the value to 'fill the gap' at a removable discontinuity?

$f(a) = \lim_{x \to a} f(x)$ , where 'a' is the x-value of the discontinuity.

Write the simplified form of a function with a removable discontinuity at x=c after factoring.

If $f(x) = \frac{(x-c)g(x)}{(x-c)}$ , then the simplified function is $g(x)$ for $x \neq c$ .

What is the condition for the existence of a limit at a point?

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$

How can you express a piecewise function generally?

$f(x) = \begin{cases} f_1(x), & x \in D_1 \\ f_2(x), & x \in D_2 \\ ... \end{cases}$

How do you find the x-value(s) where a rational function might have discontinuities?

Solve $q(x) = 0$ where $f(x) = \frac{p(x)}{q(x)}$

How to redefine f(x) to remove discontinuity at x=a?

$f(x) = \begin{cases} original, & x \neq a \\ \lim_{x \to a} original, & x = a \end{cases}$

What is the simplified function after removing the discontinuity?

$f(x) = \frac{(x-a)g(x)}{(x-a)} = g(x)$

How do you solve for constant 'k' to make piecewise function continuous?

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = k$