Removing Discontinuities - AP Calculus AB

For $r(x)=\begin{cases}\dfrac{x^2-36}{x-6},&x\ne6\\11,&x=6\end{cases}$ , what value should replace $11$ to make $r$ continuous at $x=6$ ?
Answer: $12$ r(x) features a removable discontinuity at x=6 from the common (x-6) factor. Set r(6) to the limit. Simplifies to x+6, limit 6+6=12. Replace 11 with 12, choice B. Common confusion: direct substitution yields undefined, but canceling fixes it. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
For $g(x)=\begin{cases}\dfrac{x^2-x-12}{x-4},&x\ne4\\0,&x=4\end{cases}$ , what value should replace $0$ to make $g$ continuous at $x=4$ ?
Answer: $7$ g(x) has a removable discontinuity at x=4 with shared (x-4). Set to limit: (x-4)(x+3)/(x-4) = x+3, 4+3=7. Replace 0 with 7, choice C. Confusion from direct sub: 0/0. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let $s(x)=\begin{cases}\dfrac{x^2-49}{x-7},&x\ne7\\20,&x=7\end{cases}$ . What value should replace $20$ to remove the discontinuity at $x=7$ ?
Answer: $14$ 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 $v(x)=\begin{cases}\frac{x^2-49}{x-7},&x\ne7\\12,&x=7\end{cases}$ . What value at $x=7$ makes $v$ continuous?
Answer: $14$ The function v(x) has a removable discontinuity at x = 7 because the numerator x² - 49 = (x - 7)(x + 7) contains the factor (x - 7) that cancels with the denominator. After canceling, we get v(x) = x + 7 for x ≠ 7. To remove the discontinuity, we need v(7) = lim(x→7) (x + 7) = 7 + 7 = 14. Students might keep the current value of 12 or forget to add when evaluating. The key strategy is recognizing difference of squares (a² - b² = (a-b)(a+b)), canceling common factors, then evaluating the simplified expression.
Let $r(x)=\begin{cases}\frac{x^2-25}{x-5},&x\ne5\\11,&x=5\end{cases}$ . What value at $x=5$ makes $r$ continuous?
Answer: $10$ 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.
For $f(x)=\begin{cases}\frac{x^2-9}{x-3},&x\ne3\\5,&x=3\end{cases}$ , what value should replace $f(3)$ to remove the discontinuity?
Answer: $6$ The function has a removable discontinuity at x = 3 because the numerator x² - 9 = (x - 3)(x + 3) contains the factor (x - 3) that cancels with the denominator. After canceling, we get f(x) = x + 3 for x ≠ 3. To find the value that removes the discontinuity, we calculate lim(x→3) f(x) = lim(x→3) (x + 3) = 3 + 3 = 6. A common mistake is to think the function equals 0 at the discontinuity or to use the current defined value of 5. The strategy is: factor the numerator, cancel common factors with the denominator, then evaluate the simplified expression at the point of discontinuity.
Let $h(x)=\begin{cases}\dfrac{x^2-1}{x-1},&x\ne1\\7,&x=1\end{cases}$ . What value should replace $7$ to remove the discontinuity at $x=1$ ?
Answer: $2$ 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 $h(x)=\begin{cases}\dfrac{x^2-3x+2}{x-1},&x\ne1\\8,&x=1\end{cases}$ . What value should replace $8$ to remove the discontinuity at $x=1$ ?
Answer: $-1$ h(x) shows removable discontinuity at x=1. Simplifies: (x-1)(x-2)/(x-1) = x-2, limit 1-2=-1. Replace 8 with -1, choice E. Common error: undefined without simplifying. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.
Let $q(x)=\begin{cases}\dfrac{x^2-6x+5}{x-1},&x\ne1\\0,&x=1\end{cases}$ . What value should replace $0$ to remove the discontinuity at $x=1$ ?
Answer: $-4$ 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 $r(x)=\begin{cases}\dfrac{x^2+3x-10}{x-2},&x\ne2\\9,&x=2\end{cases}$ , what value should replace $r(2)$ to remove the discontinuity?
Answer: $7$ The function r(x) has a removable discontinuity at x = 2 because we can factor the numerator as x² + 3x - 10 = (x - 2)(x + 5), revealing the common factor (x - 2) with the denominator. After canceling, we get r(x) = x + 5 for x ≠ 2. To remove the discontinuity, we must set r(2) = lim(x→2) r(x) = lim(x→2) (x + 5) = 2 + 5 = 7. The given value r(2) = 9 causes the discontinuity by being 2 units above the limit. Students often struggle with factoring trinomials where the leading coefficient is 1; the key is finding two numbers that multiply to -10 and add to 3. Always verify your factorization before canceling to ensure accuracy in finding the correct limit.
For $g(x)=\begin{cases}\dfrac{x^2-4}{x-2},&x\ne2\\7,&x=2\end{cases}$ , what value should replace $7$ to make $g$ continuous at $x=2$ ?
Answer: $4$ At x = 2, the function g(x) has a removable discontinuity because (x² - 4)/(x - 2) is undefined there. We factor the numerator as (x² - 4) = (x + 2)(x - 2), giving us (x + 2)(x - 2)/(x - 2) = x + 2 for x ≠ 2. The limit as x approaches 2 is lim[x→2] (x + 2) = 2 + 2 = 4. Therefore, to make g continuous at x = 2, we should replace 7 with 4. Students often forget that continuity requires the function value to equal the limit at that point. The strategy is to factor, simplify, then evaluate the limit of the simplified expression.
Let $p(x)=\begin{cases}\frac{x^2-1}{x+1},&x\ne-1\\2,&x=-1\end{cases}$ . What value at $x=-1$ removes the discontinuity?
Answer: $-2$ The function p(x) has a removable discontinuity at x = -1 because the numerator x² - 1 = (x + 1)(x - 1) contains the factor (x + 1) that cancels with the denominator. After canceling, we get p(x) = x - 1 for x ≠ -1. To remove the discontinuity, we need p(-1) = lim(x→-1) (x - 1) = -1 - 1 = -2. Students might mistakenly use the current value of 2 or evaluate without factoring first. Remember: removable discontinuities occur when the same factor appears in both numerator and denominator, and the limit after canceling gives the continuous value.
Let $q(x)=\begin{cases}\dfrac{x^2-25}{x-5},&x\ne5\\0,&x=5\end{cases}$ . What value should replace $0$ to remove the discontinuity at $x=5$ ?
Answer: $10$ 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.
For $f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\ne3\\5,&x=3\end{cases}$ , what value should replace $5$ to remove the discontinuity at $x=3$ ?
Answer: $6$ The function has a removable discontinuity at x = 3 because the rational expression (x² - 9)/(x - 3) is undefined there. To find the correct value, we factor the numerator: (x² - 9) = (x + 3)(x - 3), so the expression becomes (x + 3)(x - 3)/(x - 3) = x + 3 for x ≠ 3. Taking the limit as x approaches 3, we get lim[x→3] (x + 3) = 3 + 3 = 6. A common mistake is to substitute x = 3 directly into the original expression without factoring first, which would give 0/0. To remove any removable discontinuity, always find the limit of the simplified expression and set the function value equal to that limit.
For $f(x)=\dfrac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=8$ , what value should replace $f(3)$ to remove the discontinuity?
Answer: $6$ 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.
For $p(x)=\begin{cases}\dfrac{x^2-16}{x-4},&x\ne4\\10,&x=4\end{cases}$ , what value should replace $10$ to make $p$ continuous at $x=4$ ?
Answer: $8$ 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.
For $u(x)=\begin{cases}\dfrac{x^2-49}{x+7},&x\ne-7\\1,&x=-7\end{cases}$ , what value of $u(-7)$ removes the discontinuity?
Answer: $-14$ This function has a removable discontinuity at x = -7 because the numerator x² - 49 = (x - 7)(x + 7) contains the factor (x + 7) that cancels with the denominator. After canceling, we get u(x) = x - 7 for x ≠ -7. To make u continuous, we need u(-7) = lim(x→-7) u(x) = lim(x→-7) (x - 7) = -7 - 7 = -14. The current value u(-7) = 1 creates a 15-unit jump discontinuity. A common mistake is to cancel (x - 7) instead of (x + 7), or to evaluate -7 + 7 = 0 instead of -7 - 7 = -14. Remember that when the discontinuity is at a negative value, be extra careful with signs when evaluating the simplified expression.
For $q(x)=\begin{cases}\frac{x^2-16}{x-4},&x\ne4\\9,&x=4\end{cases}$ , what value should replace $q(4)$ to remove the discontinuity?
Answer: $8$ The function q(x) has a removable discontinuity at x = 4 because the numerator x² - 16 = (x - 4)(x + 4) contains the factor (x - 4) that cancels with the denominator. After canceling, we get q(x) = x + 4 for x ≠ 4. To make q continuous, we need q(4) = lim(x→4) (x + 4) = 4 + 4 = 8. A common mistake is keeping the current value of 9 or thinking the function should equal 0 at the discontinuity. The strategy is to recognize difference of squares factorizations, cancel common factors, then evaluate at the point of discontinuity.
Let $q(x)=\begin{cases}\dfrac{x^2-16}{x-4},&x\ne4\\10,&x=4\end{cases}$ . What value should replace $10$ to make $q$ continuous at $x=4$ ?
Answer: $8$ At x = 4, the function q(x) has a removable discontinuity since (x² - 16)/(x - 4) is undefined. We recognize that x² - 16 = (x + 4)(x - 4), so the expression becomes (x + 4)(x - 4)/(x - 4) = x + 4 for x ≠ 4. Taking the limit as x approaches 4, we get lim[x→4] (x + 4) = 4 + 4 = 8. Therefore, to make q continuous at x = 4, we should replace 10 with 8. A common mistake is to think the discontinuity cannot be removed, but factoring reveals it's removable. The key strategy is to factor difference of squares patterns and simplify before taking limits.
Define $p(x)=\begin{cases}\dfrac{x^2-16}{x-4},&x\ne4\\3,&x=4\end{cases}$ . What value replaces $p(4)$ to remove the discontinuity?
Answer: $8$ The function p(x) has a removable discontinuity at x = 4 because the numerator x² - 16 = (x - 4)(x + 4) contains the factor (x - 4) that cancels with the denominator. After canceling, we obtain p(x) = x + 4 for x ≠ 4. To remove the discontinuity, we must set p(4) equal to lim(x→4) p(x) = lim(x→4) (x + 4) = 4 + 4 = 8. The given value p(4) = 3 is what causes the discontinuity, as it doesn't match the limit. Students sometimes think the answer should be related to the difference 16 - 4 = 12, but this is incorrect. The strategy for removing discontinuities is consistent: factor, cancel, then evaluate the simplified expression at the point of discontinuity.

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