What does a hole in the graph of f(x) indicate?
A removable discontinuity at that x-value.
How can you visually identify a removable discontinuity on a graph?
Look for a point where the graph is undefined (an open circle or 'hole'), but the graph approaches a specific y-value from both sides.
If the graph of a function has a 'jump', what type of discontinuity is it?
A non-removable discontinuity (specifically, a jump discontinuity).
What does a vertical asymptote on the graph of a function indicate?
An infinite discontinuity at that x-value.
How does the graph of a piecewise function look when it's continuous?
The pieces of the graph connect smoothly at the boundaries, without any gaps or jumps.
How does the graph of $f(x) = \frac{x^2 - 4}{x - 2}$ look near x=2?
Like the line y = x + 2, but with a hole at the point (2, 4).
If a graph has a removable discontinuity at x = a, what does the limit as x approaches a represent graphically?
The y-value that the graph approaches as x gets closer to a, even though the function is not defined there.
What does it mean if you can trace a graph without lifting your pencil?
The function is continuous over that interval.
How can you tell if a piecewise function is continuous by looking at its graph?
The pieces of the graph must meet at the boundaries without any jumps or breaks.
What does the absence of holes, jumps, or vertical asymptotes suggest about a function's continuity?
The function is likely continuous over its domain.
What are the differences between removable and non-removable discontinuities?
Removable: Limit exists, can be 'fixed'. Non-removable: Limit doesn't exist, cannot be 'fixed'.
Compare jump and infinite discontinuities.
Jump: Function 'jumps' from one value to another. Infinite: Function approaches infinity (vertical asymptote).
What are the differences between direct substitution and factoring when evaluating limits?
Direct Substitution: Simple, works if function is continuous. Factoring: Used when direct substitution leads to indeterminate form.
Compare the graphs of continuous and discontinuous functions.
Continuous: No breaks, holes, or jumps. Discontinuous: Has breaks, holes, jumps, or asymptotes.
What are the differences between finding limits graphically and algebraically?
Graphically: Visual estimation of where the function is heading. Algebraically: Using techniques like factoring or rationalizing to find the exact value.
Compare the conditions for continuity at a point versus over an interval.
At a point: Limit exists and equals the function's value. Over an interval: Continuous at every point in the interval.
What are the differences between a hole and a vertical asymptote on a graph?
Hole: Removable discontinuity where the limit exists. Vertical Asymptote: Non-removable discontinuity where the function approaches infinity.
Compare the methods for finding discontinuities in rational functions versus piecewise functions.
Rational Functions: Look for zeros in the denominator. Piecewise Functions: Check the boundaries between the pieces.
What are the differences between the Intermediate Value Theorem and the Extreme Value Theorem?
Intermediate Value Theorem: Guarantees a function takes on every value between any two given points if continuous. Extreme Value Theorem: Guarantees the function has a max and min if continuous.
Compare direct substitution and L'Hopital's rule for evaluating limits.
Direct Substitution: First attempt, works if not indeterminate. L'Hopital's Rule: Used for indeterminate forms like 0/0 or ∞/∞, involves derivatives.
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: $f(-2) = -2 + 1 = -1$.
