All Flashcards
Define continuity on an interval.
A function is continuous on an interval if it is continuous at every point within that interval.
What is a piecewise function?
A function defined by multiple sub-functions, each applying to a certain interval of the main function's domain.
What is a domain restriction?
A limitation on the possible input values (x-values) for a function, often due to square roots or division by zero.
Define left-hand limit.
The value a function approaches as the input approaches a given value from the left side.
Define right-hand limit.
The value a function approaches as the input approaches a given value from the right side.
What does it mean for a function to be continuous at a point?
The left-hand limit, right-hand limit, and the function's value at that point must all be equal.
What are rational functions?
Functions that can be expressed as the quotient of two polynomials.
What are radical functions?
Functions containing a radical, such as a square root or cube root.
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 polynomial function?
A function that can be written in the form (f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0), where n is a non-negative integer and the coefficients are constants.
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}).