All Flashcards
Explain the concept of continuity for polynomials.
Polynomials are continuous everywhere because they have no breaks, jumps, or vertical asymptotes in their graphs.
Explain the concept of continuity for rational functions.
Rational functions are continuous everywhere except where the denominator is zero, which creates a vertical asymptote or a hole.
Explain the concept of checking continuity at a point for piecewise functions.
You must verify that the left-hand limit, right-hand limit, and the function's value at that point are all equal to ensure there are no jumps or breaks.
Why are domain restrictions important when checking continuity?
Domain restrictions indicate points where the function is not defined, leading to discontinuities. These points must be excluded when determining continuity over an interval.
Explain the relationship between limits and continuity.
For a function to be continuous at a point, the limit as x approaches that point must exist and be equal to the function's value at that point.
Explain the concept of removable discontinuity.
A removable discontinuity occurs when a function has a hole at a point, meaning the limit exists, but the function is either undefined or has a different value at that point. It can be 'removed' by redefining the function at that point.
Explain the concept of non-removable discontinuity.
A non-removable discontinuity occurs when the limit does not exist at a point, such as at a vertical asymptote or a jump discontinuity. It cannot be 'removed' by redefining the function at that point.
Explain the importance of checking both left-hand and right-hand limits.
Checking both limits is essential for determining continuity, especially in piecewise functions, because the function's behavior may differ on either side of a point.
Explain the concept of continuity for trigonometric functions.
Trigonometric functions like sine and cosine are continuous everywhere. Tangent, cotangent, secant, and cosecant are continuous on their domains, excluding points where they have vertical asymptotes.
Explain the concept of continuity for exponential and logarithmic functions.
Exponential functions are continuous everywhere. Logarithmic functions are continuous on their domains, which are typically (x > 0).
How do you determine if a piecewise function is continuous at a point where the definition changes?
Step 1: Find the left-hand limit. Step 2: Find the right-hand limit. Step 3: Evaluate the function at that point. Step 4: Compare all three values; they must be equal for continuity.
How do you check for continuity of (f(x) = \frac{p(x)}{q(x)}) over an interval?
Step 1: Identify values where (q(x) = 0). Step 2: Check if these values are within the interval. Step 3: If any are, the function is discontinuous over the interval.
How do you determine the domain of a square root function, like (f(x) = \sqrt{g(x)})?
Step 1: Set the expression inside the square root greater than or equal to zero: (g(x) \geq 0). Step 2: Solve the inequality for (x). Step 3: The solution is the domain of the function.
How do you find the value of 'k' that makes a piecewise function continuous?
Step 1: Set the pieces of the function equal to each other at the point where the definition changes. Step 2: Solve for 'k'. Step 3: Verify that this value of 'k' makes the function continuous by checking the left-hand and right-hand limits.
How do you determine if a function is continuous on an open interval?
Step 1: Check for any domain restrictions within the interval. Step 2: If there are no domain restrictions, the function is likely continuous. Step 3: For piecewise functions, check continuity at each transition point.
Given (f(x) = \frac{x^2 - 4}{x - 2}), is it continuous at (x = 2)? If not, what type of discontinuity is it?
Step 1: Simplify the function: (f(x) = x + 2) for (x \neq 2). Step 2: Check the limit as (x \to 2): (\lim_{x \to 2} (x + 2) = 4). Step 3: Since the limit exists but (f(2)) is undefined, it's a removable discontinuity.
How to show (f(x)) is continuous on ([a,b])?
Step 1: Show (f(x)) is continuous on ((a,b)). Step 2: Show (\lim_{x \to a^+} f(x) = f(a)). Step 3: Show (\lim_{x \to b^-} f(x) = f(b)).
How to find the interval(s) where (f(x) = \sqrt{x^2 - 5x + 6}) is continuous?
Step 1: Solve (x^2 - 5x + 6 \geq 0). Step 2: Factor the quadratic: ((x - 2)(x - 3) \geq 0). Step 3: Determine the intervals where the inequality holds: ((-\infty, 2] \cup [3, \infty)).
How to determine if (f(x) = \begin{cases} x^2, & x < 1 \ 2x - 1, & x \geq 1 \end{cases}) is continuous at (x = 1)?
Step 1: Find the left-hand limit: (\lim_{x \to 1^-} x^2 = 1). Step 2: Find the right-hand limit: (\lim_{x \to 1^+} (2x - 1) = 1). Step 3: Evaluate (f(1) = 2(1) - 1 = 1). Step 4: Since all three values are equal, the function is continuous at (x = 1).
How to find the value of 'c' such that (f(x) = \begin{cases} cx + 2, & x \leq 2 \ x^2 - cx, & x > 2 \end{cases}) is continuous at (x = 2)?
Step 1: Set the two pieces equal at (x = 2): (c(2) + 2 = (2)^2 - c(2)). Step 2: Solve for 'c': (2c + 2 = 4 - 2c \Rightarrow 4c = 2 \Rightarrow c = \frac{1}{2}).
How does a vertical asymptote on the graph of a function relate to its continuity?
A vertical asymptote indicates a point of infinite discontinuity, meaning the function is not continuous at that x-value.
How does a hole in the graph of a function relate to its continuity?
A hole indicates a removable discontinuity. The function is not continuous at that x-value unless redefined to fill the hole.
How does a jump in the graph of a function relate to its continuity?
A jump indicates a non-removable discontinuity. The function is not continuous at that x-value because the left-hand and right-hand limits are not equal.
What does a smooth, unbroken graph indicate about the continuity of a function?
A smooth, unbroken graph indicates that the function is continuous over the interval shown.
If a graph has a cusp (sharp point), is the function continuous at that point?
Yes, the function can be continuous, but it is not differentiable at a cusp.
How can you visually determine if a piecewise function is continuous from its graph?
Check if the different pieces of the graph connect seamlessly without any jumps, breaks, or holes at the points where the function's definition changes.
What does the graph of (f(x) = \frac{1}{x}) tell us about its continuity?
The graph has a vertical asymptote at (x = 0), indicating a non-removable discontinuity. The function is continuous on ((-\infty, 0) \cup (0, \infty)).
What does the graph of (f(x) = \sqrt{x}) tell us about its continuity?
The graph starts at ((0, 0)) and is smooth and unbroken for (x \geq 0), indicating that the function is continuous on ([0, \infty)).
How to identify a removable discontinuity from a graph?
Look for a hole (open circle) in the graph. The function is not defined at that point, or the function value does not match the limit at that point.
How to identify a jump discontinuity from a graph?
Look for a sudden jump in the graph, where the left-hand and right-hand limits are different.