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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.
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.