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  2. AP Calculus
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What does a hole in the graph of f(x) indicate?

A removable discontinuity at that x-value.

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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)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​ 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.

Explain the concept of removing a discontinuity.

It involves redefining the function at the point of discontinuity to equal the limit at that point, effectively 'filling the hole' and making the function continuous.

Explain how factoring helps in identifying removable discontinuities.

Factoring allows you to simplify rational functions and cancel out common factors that cause discontinuities, revealing the simplified function and the location of the hole.

Describe the conditions needed for a piecewise function to be continuous at a boundary.

The left-hand limit and the right-hand limit at the boundary must exist and be equal to each other, and the function's value at the boundary must also be equal to this limit.

Explain why a removable discontinuity is not a 'true' discontinuity.

Because the limit exists at that point, and by redefining the function at that point, we can make the function continuous.

How does the graph of a function with a removable discontinuity look?

The graph appears continuous except for a single point (a 'hole') where the function is undefined.

Why is it important to check both left-hand and right-hand limits for piecewise functions?

Because the function's behavior may be different on either side of the boundary point, and continuity requires these limits to match.

What are the key steps to determine if a rational function has a removable discontinuity?

Factor the numerator and denominator, cancel any common factors, and identify the x-values that were canceled out.

Explain how the concept of a limit is essential for understanding removable discontinuities.

The limit tells us where the function 'should' be, even if it isn't defined there. This allows us to redefine the function and remove the discontinuity.

How does visualizing a graph help in identifying discontinuities?

Graphs provide a visual representation of where a function might have breaks, jumps, or holes, making it easier to spot potential discontinuities.

Explain why setting the numerator and denominator equal to zero is important.

Setting the denominator equal to zero helps to find potential points of discontinuity, and setting the numerator equal to zero helps to find the x-intercepts of the function.

How to find and remove a discontinuity in f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}f(x)=x−2x2−4​?

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

How to ensure continuity of f(x)={x2,x≤1ax,x>1f(x) = \begin{cases} x^2, & x \leq 1 \\ ax, & x > 1 \end{cases}f(x)={x2,ax,​x≤1x>1​?

Set x2=axx^2 = axx2=ax at x=1x = 1x=1. Solve for aaa: 12=a(1)1^2 = a(1)12=a(1), so a=1a = 1a=1.

How to determine if f(x)=x2−1x+1f(x) = \frac{x^2 - 1}{x + 1}f(x)=x+1x2−1​ has a removable discontinuity?

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

How to redefine f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}f(x)=x−3x2−9​ to be continuous?

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

How to find the value of 'a' to make f(x)={x+a,x<0x2,x≥0f(x) = \begin{cases} x + a, & x < 0 \\ x^2, & x \geq 0 \end{cases}f(x)={x+a,x2,​x<0x≥0​ continuous?

Set x+a=x2x + a = x^2x+a=x2 at x=0x = 0x=0. Solve for aaa: 0+a=020 + a = 0^20+a=02, so a=0a = 0a=0.

Given f(x)=x2−4x+3x−1f(x) = \frac{x^2 - 4x + 3}{x - 1}f(x)=x−1x2−4x+3​, how do you find and remove the discontinuity?

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

How do you ensure the function f(x)={cx2,x≤22x+c,x>2f(x) = \begin{cases} cx^2, & x \leq 2 \\ 2x + c, & x > 2 \end{cases}f(x)={cx2,2x+c,​x≤2x>2​ is continuous?

Set the two pieces equal at x = 2: c(2)2=2(2)+cc(2)^2 = 2(2) + cc(2)2=2(2)+c. Solve for c: 4c=4+c4c = 4 + c4c=4+c, so 3c=43c = 43c=4, and c=43c = \frac{4}{3}c=34​.

How do you find the limit of f(x)=x2−25x+5f(x) = \frac{x^2 - 25}{x + 5}f(x)=x+5x2−25​ as x approaches -5?

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

How do you find the value of 'k' that makes f(x)={x2−9x−3,x≠3k,x=3f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases}f(x)={x−3x2−9​,k,​x=3x=3​ continuous?

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

How do you determine if the function f(x)=x2+3x+2x+2f(x) = \frac{x^2 + 3x + 2}{x + 2}f(x)=x+2x2+3x+2​ has a removable discontinuity and remove it?

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