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  2. AP Calculus
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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.

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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.

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.

How do you test continuity for a piecewise function at x=a?

lim⁡x→a−f(x)=lim⁡x→a+f(x)=f(a)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)limx→a−​f(x)=limx→a+​f(x)=f(a)

What is the general form of a rational function where removable discontinuities often occur?

f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)}f(x)=q(x)p(x)​, where p(x) and q(x) are polynomials.

How do you find the value to 'fill the gap' at a removable discontinuity?

f(a)=lim⁡x→af(x)f(a) = \lim_{x \to a} f(x)f(a)=limx→a​f(x), where 'a' is the x-value of the discontinuity.

Write the simplified form of a function with a removable discontinuity at x=c after factoring.

If f(x)=(x−c)g(x)(x−c)f(x) = \frac{(x-c)g(x)}{(x-c)}f(x)=(x−c)(x−c)g(x)​, then the simplified function is g(x)g(x)g(x) for x≠cx \neq cx=c.

What is the condition for the existence of a limit at a point?

lim⁡x→a−f(x)=lim⁡x→a+f(x)\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)limx→a−​f(x)=limx→a+​f(x)

How can you express a piecewise function generally?

f(x)={f1(x),x∈D1f2(x),x∈D2...f(x) = \begin{cases} f_1(x), & x \in D_1 \\ f_2(x), & x \in D_2 \\ ... \end{cases}f(x)=⎩⎨⎧​f1​(x),f2​(x),...​x∈D1​x∈D2​​

How do you find the x-value(s) where a rational function might have discontinuities?

Solve q(x)=0q(x) = 0q(x)=0 where f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)}f(x)=q(x)p(x)​

How to redefine f(x) to remove discontinuity at x=a?

f(x)={original,x≠alim⁡x→aoriginal,x=af(x) = \begin{cases} original, & x \neq a \\ \lim_{x \to a} original, & x = a \end{cases}f(x)={original,limx→a​original,​x=ax=a​

What is the simplified function after removing the discontinuity?

f(x)=(x−a)g(x)(x−a)=g(x)f(x) = \frac{(x-a)g(x)}{(x-a)} = g(x)f(x)=(x−a)(x−a)g(x)​=g(x)

How do you solve for constant 'k' to make piecewise function continuous?

lim⁡x→a−f(x)=lim⁡x→a+f(x)=k\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = klimx→a−​f(x)=limx→a+​f(x)=k