Removing Discontinuities - AP Calculus BC

A function is $g(x)=\frac{x^2-4}{x+2}$ for $x\ne-2$ and $g(-2)=3$ ; what value makes $g$ continuous at $x=-2$ ?
Answer: $-4$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=-2, we need to set g(-2) equal to the limit as x approaches -2 of g(x). By factoring the numerator as (x-2)(x+2), we can cancel the (x+2) term with the denominator, simplifying to x-2 for x ≠ -2. Therefore, the limit as x approaches -2 is -2-2=-4, which is the value that removes the discontinuity. A tempting distractor like 3 might be the originally given value of g(-2), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A function is $v(x)=\frac{x^2-100}{x-10}$ for $x\ne10$ and $v(10)=5$ ; what value removes the discontinuity at $x=10$ ?
Answer: $20$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=10, we need to set v(10) equal to the limit as x approaches 10 of v(x). By factoring the numerator as (x-10)(x+10), we can cancel the (x-10) term with the denominator, simplifying to x+10 for x ≠ 10. Therefore, the limit as x approaches 10 is 10+10=20, which is the value that removes the discontinuity. A tempting distractor like 5 might be the originally given value of v(10), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A function is $C(x)=\frac{x^2-6x}{x}$ for $x\ne0$ and $C(0)=-1$ ; what value removes the discontinuity at $x=0$ ?
Answer: $-6$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=0, we need to set C(0) equal to the limit as x approaches 0 of C(x). By factoring the numerator as x(x-6), we can cancel the x term with the denominator, simplifying to x-6 for x ≠ 0. Therefore, the limit as x approaches 0 is 0-6=-6, which is the value that removes the discontinuity. A tempting distractor like -1 might be the originally given value of C(0), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A particle's position uses $p(t)=\frac{t^2-16}{t-4}$ for $t\ne4$ and $p(4)=1$ ; what value makes $p$ continuous at $t=4$ ?
Answer: $8$ This problem focuses on removing a removable discontinuity in a rational function. The function p(t) = (t² - 16)/(t - 4) has a discontinuity at t=4, but factoring (t-4)(t+4)/(t-4) simplifies to p(t) = t+4 for t ≠ 4. The limit as t approaches 4 is 4 + 4 = 8. Setting p(4) = 8 removes the discontinuity. A tempting distractor is 1, the original value, but it doesn't equal the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A response curve is $r(x)=\frac{x^2+5x}{x}$ for $x \ne 0$ and $r(0)=0$ ; what value of $r(0)$ removes the discontinuity?
Answer: $5$ This problem focuses on removing a removable discontinuity in a rational function. The function $r(x) = \frac{x^2 + 5x}{x}$ simplifies to $x + 5$ for $x \ne 0$ after canceling x. The limit as $x$ approaches 0 is $0 + 5 = 5$ . Defining $r(0) = 5$ removes the discontinuity. A tempting distractor is 0, the original value, but it doesn't match the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A function is $p(x)=\frac{x^2-14x+49}{x-7}$ for $x\ne7$ and $p(7)=3$ ; what value removes the discontinuity at $x=7$ ?
Answer: $0$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=7, we need to set p(7) equal to the limit as x approaches 7 of p(x). By factoring the numerator as (x-7)(x-7), we can cancel one (x-7) term with the denominator, simplifying to x-7 for x ≠ 7. Therefore, the limit as x approaches 7 is 7-7=0, which is the value that removes the discontinuity. A tempting distractor like 3 might be the originally given value of p(7), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
Define $v(x)=\frac{x^2+x}{x}$ for $x\ne0$ and $v(0)=9$ . What value of $v(0)$ removes the discontinuity?
Answer: $1$ This question involves removing a removable discontinuity at x = 0. To make v continuous, v(0) must equal the limit as x approaches 0. We can factor out x from the numerator: v(x) = x(x+1)/x = x+1 for x≠0. Therefore, lim[x→0] v(x) = 0+1 = 1, so v(0) should be 1 to remove the discontinuity. The given value v(0) = 9 is much larger than the limit and maintains the discontinuity. When a rational function has x as a factor in both numerator and denominator, always factor it out first before evaluating the limit.
A controller uses $k(x)=\frac{x^2-6x+9}{x-3}$ for $x \neq 3$ and $k(3)=2$ ; what value of $k(3)$ removes the discontinuity?
Answer: $0$ This problem focuses on removing a removable discontinuity in a rational function. The function $k(x) = \frac{x^2 - 6x + 9}{x - 3}$ factors to $\frac{(x-3)^2}{x-3}$ , simplifying to $x - 3$ for $x \neq 3$ . The limit as x approaches 3 is $3 - 3 = 0$ . Setting k(3) = 0 removes the discontinuity. A tempting distractor is 2, the original value, but it doesn't equal the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A function is $f(x)=\frac{x^2-2x}{x}$ for $x\ne0$ with $f(0)=4$ ; what value of $f(0)$ removes the discontinuity?
Answer: $-2$ This problem focuses on removing a removable discontinuity in a rational function. The function f(x) = (x² - 2x)/x simplifies to x - 2 for x ≠ 0 after canceling x. The limit as x approaches 0 is 0 - 2 = -2. Setting f(0) = -2 removes the discontinuity. A tempting distractor is 4, the original value, but it doesn't match the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A function is $n(x)=\frac{x^2-5x}{x}$ for $x\ne0$ and $n(0)=5$ ; what value makes $n$ continuous at $x=0$ ?
Answer: $-5$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=0, we need to set n(0) equal to the limit as x approaches 0 of n(x). By factoring the numerator as x(x-5), we can cancel the x term with the denominator, simplifying to x-5 for x ≠ 0. Therefore, the limit as x approaches 0 is 0-5=-5, which is the value that removes the discontinuity. A tempting distractor like 5 might come from ignoring the sign in the simplified expression. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A lab defines $f(x)=\frac{(x-2)(x+2)}{x-2}$ for $x\ne2$ and $f(2)=-1$ ; what value should $f(2)$ be to remove the discontinuity?
Answer: $4$ This problem focuses on removing a removable discontinuity in a rational function. The function f(x) = ((x-2)(x+2))/(x-2) simplifies to x + 2 for x ≠ 2 after canceling (x-2). The limit as x approaches 2 is 2 + 2 = 4. Setting f(2) = 4 removes the discontinuity. A tempting distractor is -1, the original value, but it ignores the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
An index is $I(x)=\frac{x^2-3x}{x}$ for $x\ne0$ and $I(0)=0$ ; what value makes $I$ continuous at $x=0$ ?
Answer: $-3$ This problem focuses on removing a removable discontinuity in a rational function. The function $I(x) = \frac{x^2 - 3x}{x}$ simplifies to $x - 3$ for $x \ne 0$ after canceling x. The limit as $x$ approaches 0 is $0 - 3 = -3$ . Setting $I(0) = -3$ removes the discontinuity. A tempting distractor is $0$ , the original value, but it doesn't match the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A function is $w(x)=\frac{x^3-1}{x-1}$ for $x \neq 1$ and $w(1)=2$ ; what value makes $w$ continuous at $x=1$ ?
Answer: $3$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at $x=1$ , we need to set $w(1)$ equal to the limit as $x$ approaches 1 of $w(x)$ . By factoring the numerator as $(x-1)(x^2 + x + 1)$ , we can cancel the $(x-1)$ term with the denominator, simplifying to $x^2 + x + 1$ for $x \neq 1$ . Therefore, the limit as $x$ approaches 1 is $1+1+1=3$ , which is the value that removes the discontinuity. A tempting distractor like 2 might be the originally given value of $w(1)$ , but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A function is $B(x)=\frac{x^2-64}{x+8}$ for $x\ne-8$ and $B(-8)=1$ ; what value makes $B$ continuous at $x=-8$ ?
Answer: $-16$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=-8, we need to set B(-8) equal to the limit as x approaches -8 of B(x). By factoring the numerator as (x-8)(x+8), we can cancel the (x+8) term with the denominator, simplifying to x-8 for x ≠ -8. Therefore, the limit as x approaches -8 is -8-8=-16, which is the value that removes the discontinuity. A tempting distractor like 1 might be the originally given value of B(-8), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A function is $m(x)=\frac{x^2-81}{x-9}$ for $x\ne9$ and $m(9)=0$ ; what value removes the discontinuity at $x=9$ ?
Answer: $18$ This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=9, we need to set m(9) equal to the limit as x approaches 9 of m(x). By factoring the numerator as (x-9)(x+9), we can cancel the (x-9) term with the denominator, simplifying to x+9 for x ≠ 9. Therefore, the limit as x approaches 9 is 9+9=18, which is the value that removes the discontinuity. A tempting distractor like 0 might be the originally given value of m(9), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.
A quantity is defined by $s(x)=\frac{x^2-25}{x+5}$ for $x\ne-5$ and $s(-5)=1$ . What value of $s(-5)$ removes the discontinuity?
Answer: $-10$ To remove the removable discontinuity at x = -5, we need s(-5) to equal the limit. The function s(x) = (x²-25)/(x+5) has x²-25 = (x-5)(x+5) in the numerator, so s(x) = (x-5)(x+5)/(x+5) = x-5 for x≠-5. Therefore, lim[x→-5] s(x) = -5-5 = -10, meaning s(-5) should be -10 to make the function continuous. The given value s(-5) = 1 creates a jump discontinuity rather than removing the removable one. When dealing with difference of squares in the numerator, one factor will always cancel with the denominator at the point of discontinuity.
For a cost function $h(x)=\frac{x^2-4x}{x}$ for $x\ne0$ and $h(0)=7$ , what value of $h(0)$ removes the discontinuity?
Answer: $-4$ This problem focuses on removing a removable discontinuity in a rational function. The function h(x) = (x² - 4x)/x has a discontinuity at x=0, but simplifying by factoring x(x-4)/x yields h(x) = x-4 for x ≠ 0. The limit as x approaches 0 is 0 - 4 = -4. Defining h(0) = -4 removes the discontinuity. A tempting distractor is 4, perhaps from ignoring the sign or evaluating |x-4|, but the correct simplified expression gives -4. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A model defines $g(x)=\frac{x^2-1}{x-1}$ for $x \neq 1$ with $g(1)=5$ ; what value should replace $g(1)$ to remove the discontinuity?
Answer: $2$ This problem focuses on removing a removable discontinuity in a rational function. The function $g(x) = \frac{x^2 - 1}{x - 1}$ has a discontinuity at $x=1$ , but factoring the numerator as $(x-1)(x+1)$ allows canceling the $(x-1)$ term, simplifying to $g(x) = x+1$ for $x \neq 1$ . The limit as $x$ approaches 1 is $1 + 1 = 2$ . Setting $g(1) = 2$ removes the discontinuity by aligning the function value with the limit. A tempting distractor is 5, the original assigned value, but it doesn't match the limit and leaves the hole. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A sensor's calibration uses $f(x)=\frac{x^2-9}{x-3}$ for $x\ne3$ and $f(3)=0$ ; what value of $f(3)$ removes the discontinuity?
Answer: $6$ This problem focuses on removing a removable discontinuity in a rational function. The function f(x) = (x² - 9)/(x - 3) has a discontinuity at x=3 because the denominator is zero, but we can simplify by factoring the numerator as (x-3)(x+3) and canceling the common (x-3) factor, resulting in f(x) = x+3 for x ≠ 3. The limit as x approaches 3 is then 3 + 3 = 6. To remove the discontinuity, define f(3) = 6 to match this limit. A tempting distractor is 9, which comes from evaluating the numerator at x=3 without simplification, but it ignores the canceled factor creating the hole. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.
A statistic is $s(x)=\frac{x^2-10x+25}{x-5}$ for $x\ne5$ and $s(5)=1$ ; what value removes the discontinuity at $x=5$ ?
Answer: $0$ This problem focuses on removing a removable discontinuity in a rational function. The function s(x) = (x² - 10x + 25)/(x - 5) factors to (x-5)²/(x-5), simplifying to x - 5 for x ≠ 5. The limit as x approaches 5 is 5 - 5 = 0. Setting s(5) = 0 removes the discontinuity. A tempting distractor is 1, the original value, but it doesn't equal the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.

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