For , what value should replace to make continuous at ?
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AP Calculus AB Quiz
Practice Removing Discontinuities in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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For p(x)=⎩⎨⎧x−2x2−5x+6,1,x=2x=2, what value should replace 1 to make p continuous at x=2?
This quiz focuses on Removing Discontinuities, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
For p(x)=⎩⎨⎧x−2x2−5x+6,1,x=2x=2, what value should replace 1 to make p continuous at x=2?
Explanation: p(x) has removable discontinuity at x=2. Factoring: (x-2)(x-3)/(x-2) = x-3, limit 2-3=-1. Replace 1 with -1, choice A. Mistake: direct plug. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
For p(x)=⎩⎨⎧x−4x2−16,10,x=4x=4, what value should replace 10 to make p continuous at x=4?
Explanation: p(x) shows a removable discontinuity at x=4 as (x²-16) and (x-4) share (x-4). Set p(4) to the limit for continuity. Simplifying to (x-4)(x+4)/(x-4) = x+4. Limit at 4 is 4+4=8. Replace 10 with 8, choice A. Confusion arises from unsimplified substitution, giving undefined, but factoring resolves it. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let s(x)=⎩⎨⎧x−7x2−49,20,x=7x=7. What value should replace 20 to remove the discontinuity at x=7?
Explanation: s(x) has a removable discontinuity at x=7 with shared (x-7). Continuity needs limit value. Simplifies to x+7, limit 7+7=14. Replace 20 with 14, choice A. Mistake: plugging without simplifying gives 0/0. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let q(x)=⎩⎨⎧x−6x2−2x−24,0,x=6x=6. What value should replace 0 to remove the discontinuity at x=6?
Explanation: For q(x), the discontinuity at x=6 is removable because x² - 2x - 24 factors to (x-6)(x+4), canceling to x+4. The limit at x=6 is 6+4 = 10, the value to use for continuity. This fills the gap seamlessly. A common confusion is choosing -4 or another coefficient incorrectly. Some overlook the positive sign in the factor. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.
Let q(x)=⎩⎨⎧x−5x2−25,0,x=5x=5. What value should replace 0 to remove the discontinuity at x=5?
Explanation: q(x) has a removable discontinuity at x=5 due to the shared factor (x-5) in numerator and denominator. For continuity, q(5) should be the limit. Simplifies to x+5, limit 5+5=10. Replace 0 with 10, choice C. Direct plug-in confuses by giving 0/0, but simplification shows 10. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let r(x)={x−5x2−25,11,x=5x=5. What value at x=5 makes r continuous?
Explanation: The function r(x) has a removable discontinuity at x = 5 because the numerator x² - 25 = (x - 5)(x + 5) contains the factor (x - 5) that cancels with the denominator. After canceling, we get r(x) = x + 5 for x ≠ 5. To remove the discontinuity, we need r(5) = lim(x→5) (x + 5) = 5 + 5 = 10. Students often confuse this with the current value of 11 or try to evaluate the original fraction directly. The key insight is that removable discontinuities can be fixed by finding the limit of the simplified function after canceling common factors.
Let s(x)=⎩⎨⎧x−3x2−10x+21,0,x=3x=3. What value should replace 0 to remove the discontinuity at x=3?
Explanation: For s(x), there is a removable discontinuity at x=3 as the numerator x² - 10x + 21 factors into (x-3)(x-7), allowing cancellation with x-3. This simplifies to x-7 for x ≠ 3. To make s continuous, set s(3) to the limit of x-7 as x approaches 3, yielding 3-7 = -4. This fills the gap, ensuring no break in the graph. A common confusion arises from directly substituting x=3 into the unsimplified form, which is undefined, instead of simplifying. Some mix up the signs when evaluating the limit. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.
Let q(x)=⎩⎨⎧x−1x2−6x+5,0,x=1x=1. What value should replace 0 to remove the discontinuity at x=1?
Explanation: The function q(x) exhibits a removable discontinuity at x=1 since the numerator x² - 6x + 5 factors into (x-1)(x-5), sharing a common factor with the denominator x-1. Simplifying yields x-5 for x ≠ 1. To remove the discontinuity, define q(1) as the limit of x-5 as x approaches 1, which is -4. This makes the function continuous at x=1 by filling the hole in the graph. A common confusion is attempting to plug x=1 directly into the original expression, resulting in an undefined value, rather than simplifying first. Another mistake is confusing the value with the roots of the numerator. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.
For r(x)=⎩⎨⎧x+5x2−25,9,x=−5x=−5, what value should replace 9 to remove the discontinuity at x=−5?
Explanation: The function r(x) has a removable discontinuity at x = -5 where (x² - 25)/(x + 5) is undefined. Factoring the numerator as a difference of squares gives x² - 25 = (x + 5)(x - 5), so the expression simplifies to (x + 5)(x - 5)/(x + 5) = x - 5 for x ≠ -5. The limit as x approaches -5 is lim[x→-5] (x - 5) = -5 - 5 = -10. To remove the discontinuity, we must replace 9 with -10. Students often make sign errors when substituting negative values. Always factor completely, cancel common factors, then carefully evaluate the limit at the point of discontinuity.
For f(x)=x−3x2−9 when x=3 and f(3)=8, what value should replace f(3) to remove the discontinuity?
Explanation: The function f(x) = (x² - 9)/(x - 3) has a removable discontinuity at x = 3 because the numerator factors as (x - 3)(x + 3), allowing us to cancel the (x - 3) term. After cancellation, we get f(x) = x + 3 for x ≠ 3. To find the value that removes the discontinuity, we calculate the limit as x approaches 3: lim[x→3] (x + 3) = 3 + 3 = 6. The current value f(3) = 8 creates a jump discontinuity, but defining f(3) = 6 would make the function continuous. A common mistake is thinking the discontinuity cannot be removed because division by zero seems problematic, but factoring reveals the cancellation possibility. The strategy is: factor the numerator, cancel common factors with the denominator, then evaluate the simplified expression at the point of discontinuity.
Define t(x)=⎩⎨⎧x−5x2−2x−15,4,x=5x=5. What value should replace t(5) to make t continuous?
Explanation: The function t(x) has a removable discontinuity at x = 5 because the numerator x² - 2x - 15 = (x - 5)(x + 3) shares the factor (x - 5) with the denominator. Canceling this factor gives t(x) = x + 3 for x ≠ 5. To remove the discontinuity, we must set t(5) = lim(x→5) t(x) = lim(x→5) (x + 3) = 5 + 3 = 8. The given value t(5) = 4 causes a 4-unit jump at x = 5. Students sometimes confuse the factorization; to factor x² - 2x - 15, find two numbers that multiply to -15 and add to -2, which are -5 and 3. The key insight is that removable discontinuities can always be "filled in" by using the limit value at the point of discontinuity.
For t(x)=⎩⎨⎧xx2+3x,−1,x=0x=0, what value should replace −1 to remove the discontinuity at x=0?
Explanation: The function t(x) has a removable discontinuity at x = 0 where (x² + 3x)/x is undefined. We can factor out x from the numerator to get x(x + 3)/x = x + 3 for x ≠ 0. The limit as x approaches 0 is lim[x→0] (x + 3) = 0 + 3 = 3. To remove the discontinuity, we must replace -1 with 3. Students sometimes think division by zero always creates non-removable discontinuities, but when the numerator also has the same factor, it can be canceled. Always factor out common terms before evaluating limits to identify removable discontinuities.
For g(x)=⎩⎨⎧x−5x2+2x−35,0,x=5x=5, what value should replace 0 to make g continuous at x=5?
Explanation: g(x) has removable discontinuity at x=5. Factoring: (x−5)(x+7)/(x−5)=x+7, limit 5+7=12. Replace 0 with 12, choice D. Common error: unsimplified. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let h(x)=⎩⎨⎧x−1x2−1,7,x=1x=1. What value should replace 7 to remove the discontinuity at x=1?
Explanation: h(x) has a removable discontinuity at x=1 because (x²-1) and (x-1) share the factor (x-1). Continuity requires h(1) to match the limit as x approaches 1. Factoring gives (x-1)(x+1)/(x-1) = x+1 for x≠1. The limit is 1+1=2. Replace 7 with 2 for continuity, which is choice C. A common error is direct substitution, resulting in 0/0, but simplification clarifies the limit. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let u(x)=⎩⎨⎧x−9x2−81,100,x=9x=9. What value should replace 100 to remove the discontinuity at x=9?
Explanation: u(x) has removable discontinuity at x=9. Limit: x+9, 9+9=18. Replace 100 with 18, choice A. Common error: 0/0 without simplifying. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
For g(x)=⎩⎨⎧x−3x2−11x+24,10,x=3x=3, what value should replace 10 to make g continuous at x=3?
Explanation: The function g(x) has a removable discontinuity at x=3 as x² - 11x + 24 factors into (x-3)(x-8), simplifying to x-8. Set g(3) to 3-8 = -5 for continuity. This aligns the point. A common confusion is selecting a positive value or root. Undefined direct substitution misleads. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.
For t(x)=⎩⎨⎧x−8x2−64,0,x=8x=8, what value should replace 0 to make t continuous at x=8?
Explanation: t(x) shows removable discontinuity at x=8 via (x-8). Set to limit: simplifies to x+8, 8+8=16. Replace 0 with 16, choice B. Confusion from direct sub: undefined. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let f(x)=⎩⎨⎧x−8x2−18x+80,4,x=8x=8. What value should replace 4 to remove the discontinuity at x=8?
Explanation: For f(x), the discontinuity at x=8 is removable since x² - 18x + 80 factors as (x-8)(x-10), canceling with x-8 to give x-10. The limit at x=8 is 8-10 = -2, the value needed for continuity. This matches the function's behavior around the point. A common confusion is choosing a positive equivalent or ignoring the sign. Plugging in without simplification causes indefinite results. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.
For g(x)=⎩⎨⎧x−2x2−4,1,x=2x=2, what value should replace 1 to make g continuous at x=2?
Explanation: The function g(x) exhibits a removable discontinuity at x=2 since the numerator x2−4 and denominator x−2 share the factor (x−2). To make it continuous, set g(2) to the limit as x approaches 2. Simplifying, x−2(x−2)(x+2)=x+2 for x=2. The limit is 2+2=4. Replacing 1 with 4 ensures continuity, corresponding to choice D. Students often mistakenly plug in x=2 without simplifying, getting undefined, but canceling reveals the limit. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let u(x)=⎩⎨⎧x−5x2−14x+45,1,x=5x=5. What value should replace 1 to remove the discontinuity at x=5?
Explanation: In u(x), a removable discontinuity occurs at x=5 because the numerator x² - 14x + 45 factors to (x-5)(x-9), canceling with x-5. The result is x-9 for x ≠ 5. Set u(5) to the limit as x approaches 5 of x-9, equaling 5-9 = -4, to remove it. This ensures the function is defined continuously. A common confusion is assuming the value is positive or equating it to a root like 5. Direct substitution without factoring leads to errors. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.