An internet service provider charges a monthly fee based on data usage, $x$, in gigabytes. The cost function is $C(x) = \begin{cases} 30 & \text{if } 0 \le x \le 50 \\ 30 + 0.50(x-50) & \text{if } x > 50 \end{cases}$. Which statement best describes the continuity of this cost function at the $x=50$ GB threshold?
Answer: It is continuous because the cost for using slightly more than 50 GB smoothly continues from the cost of using 50 GB.
When analyzing piecewise functions, you need to check continuity by examining whether the function value approaches the same limit from both sides at the boundary point.
To test continuity at x=50, evaluate three things: the left-hand limit, right-hand limit, and the actual function value at x=50.
From the left (x→50−): Using the first piece, C(50)=30
From the right (x→50+): Using the second piece, C(x)=30+0.50(x−50). As x approaches 50, this becomes 30+0.50(50−50)=30+0=30
At the point: C(50)=30 (using the first piece since 50≤50)
Since all three values equal 30, the function is continuous at x=50. Choice A correctly identifies this continuity—the cost transitions smoothly without any sudden jumps.
Choice B is wrong because there's no removable discontinuity here; the function is already continuous. Choice C misunderstands the concept—while the rate of change (derivative) changes from 0 to 0.50, this creates a corner, not a jump discontinuity. The function value itself doesn't jump. Choice D is incorrect because the function grows linearly for x>50, so costs increase predictably, not unboundedly.
Study tip: Don't confuse discontinuity in the function with discontinuity in its derivative. A piecewise function can be continuous even when its rate of change suddenly shifts—focus on whether the function values themselves connect smoothly at boundary points.
The function $f(x) = \frac{x^2 - 1}{|x-1|}$ represents the efficiency rating of a production process. What type of discontinuity does $f(x)$ have at $x=1$?
Answer: A jump discontinuity because the left-hand and right-hand limits are finite but not equal.
To determine the type of discontinuity at $x=1$, we must evaluate the one-sided limits. The absolute value function $|x-1|$ is defined differently for $x>1$ and $x<1$.
1. Right-hand limit ($x \to 1^+$): For $x>1$, $|x-1| = x-1$. So, $\lim_{x \to 1^+} \frac{x^2 - 1}{x-1} = \lim_{x \to 1^+} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1^+} (x+1) = 2$.
2. Left-hand limit ($x \to 1^-$): For $x<1$, $|x-1| = -(x-1)$. So, $\lim_{x \to 1^-} \frac{x^2 - 1}{-(x-1)} = \lim_{x \to 1^-} \frac{(x-1)(x+1)}{-(x-1)} = \lim_{x \to 1^-} -(x+1) = -2$.
Since the left-hand limit ($-2$) and the right-hand limit ($2$) both exist but are not equal, the function has a jump discontinuity at $x=1$.
A cost function C(x)=x+3x2−9+x−12 has discontinuities at multiple points. Which statement correctly characterizes these discontinuities?
Answer: The point x=−3 has a removable discontinuity while x=1 has an infinite discontinuity
When analyzing discontinuities in rational functions, you need to examine what happens at points where the function is undefined and determine whether the discontinuity can be "fixed" through algebraic simplification.
Let's examine each discontinuity in C(x)=x+3x2−9+x−12:
For x=−3: The first fraction becomes 00, an indeterminate form. Factor the numerator: x2−9=(x−3)(x+3). This gives us x+3(x−3)(x+3)=x−3 after canceling the common factor. As x→−3, this approaches −3−3=−6. The second fraction x−12 approaches −42=−21. Since both parts have finite limits, this is a removable discontinuity.
For x=1: The first fraction x+3x2−9 approaches 4−8=−2. However, the second fraction x−12 approaches ±∞ (the sign depends on the direction of approach). This makes the entire function approach infinity, creating an infinite discontinuity.
Answer D correctly identifies this pattern. Answer A wrongly claims both are infinite discontinuities. Answer B reverses the types of discontinuities. Answer C incorrectly suggests both are removable, missing that x=1 creates an unbounded behavior.
Remember: A discontinuity is removable only if the limit exists and is finite at that point. If any part of the function approaches infinity, you have an infinite discontinuity.
Consider the function f(x)={x−2x2−45if x=2if x=2. A business analyst claims this function has a jump discontinuity at x=2. What is the most accurate assessment of this claim?
Answer: The claim is incorrect because the function has a removable discontinuity, since the limit exists but differs from the function value
First, we find the limit: limx→2x−2x2−4=limx→2x−2(x−2)(x+2)=limx→2(x+2)=4. Since limx→2f(x)=4 exists but f(2)=5=4, this is a removable discontinuity, not a jump discontinuity. Choice B is wrong because both one-sided limits equal 4. Choice C is wrong because there's no infinite discontinuity. Choice D is wrong because there's no vertical asymptote.
A revenue function R(x)=x−5∣x−5∣+2 is undefined at x=5. What happens if we define R(5)=3 to make the function continuous?
Answer: The function remains discontinuous because the left and right limits are different, creating a jump discontinuity
When analyzing continuity at a point where a function is undefined, you need to examine what happens as you approach that point from both sides and determine if assigning a value can create continuity.
Let's examine the behavior of R(x)=x−5∣x−5∣+2 near x=5. The absolute value expression ∣x−5∣ behaves differently on each side of x=5. When x<5, we have x−5<0, so ∣x−5∣=−(x−5). Therefore, x−5∣x−5∣=x−5−(x−5)=−1. When x>5, we have x−5>0, so ∣x−5∣=x−5, giving us x−5∣x−5∣=x−5x−5=1.
This means the left-hand limit is limx→5−R(x)=−1+2=1, while the right-hand limit is limx→5+R(x)=1+2=3. Since these one-sided limits are different, no two-sided limit exists.
Answer A is wrong because there is no two-sided limit to match. Answer B incorrectly suggests that undefined points can never be made continuous by assignment—they can when removable discontinuities exist. Answer C is wrong because you cannot arbitrarily assign any value to create continuity; the assigned value must equal the two-sided limit. Answer D correctly identifies that the different left and right limits create a jump discontinuity that cannot be eliminated.
Remember: A function can only be made continuous at a point by assignment if the two-sided limit exists there.
The cost, in dollars, to remove $p$ percent of a pollutant from a factory's emissions is given by the function $C(p) = \frac{75000p}{100-p}$ for $0 \le p < 100$. Analyze the behavior of this cost function as $p$ approaches 100.
Answer: The function has an infinite discontinuity at $p=100$, implying the cost becomes prohibitively large as removal approaches 100%.
We need to analyze the limit of the cost function $C(p)$ as $p$ approaches 100 from the left side (since $p<100$).
$\lim_{p \to 100^-} C(p) = \lim_{p \to 100^-} \frac{75000p}{100-p}$.
As $p$ approaches 100, the numerator approaches $75000(100) = 7,500,000$.
The denominator $(100-p)$ approaches 0 through positive values.
A non-zero number divided by a quantity approaching zero results in a limit of $\infty$ or $-\infty$. In this case, since both numerator and denominator are positive, the limit is $\infty$. This signifies an infinite discontinuity, which in this business context means the cost to remove every last particle of the pollutant would be infinitely large.
The average profit function for a new product is given by $A(x) = \frac{x^2 - 16}{x^2 - kx + 8}$, where $x$ is the number of units sold in thousands. If the function has a removable discontinuity at $x=4$, what is the value of $k$?
Answer: k=6
A rational function has a removable discontinuity at $x=a$ if both the numerator and the denominator are zero at $x=a$.
1. Check the numerator at $x=4$: $4^2 - 16 = 16 - 16 = 0$. The condition is met for the numerator.
2. For the discontinuity to be removable, the denominator must also be zero at $x=4$. Set the denominator equal to zero: $x^2 - kx + 8 = 0$.
3. Substitute $x=4$ into the expression: $4^2 - k(4) + 8 = 0$.
4. Solve for $k$: $16 - 4k + 8 = 0 \implies 24 - 4k = 0 \implies 4k = 24 \implies k = 6$. When $k=6$, the factor $(x-4)$ can be cancelled from the numerator and denominator.
A company's manufacturing cost, $C(x)$, for producing $x$ items is given by the piecewise function: C(x)={15x+50012x+kif 0≤x≤100if x>100 For the function to be continuous at $x=100$, which would prevent a sudden jump in cost at that production level, what must be the value of the constant $k$?
Answer: The value of $k$ must be 800.
For the cost function $C(x)$ to be continuous at $x=100$, the left-hand limit must equal the right-hand limit.
1. Calculate the left-hand limit: $\lim_{x \to 100^-} C(x) = \lim_{x \to 100^-} (15x + 500) = 15(100) + 500 = 1500 + 500 = 2000$.
2. Calculate the right-hand limit: $\lim_{x \to 100^+} C(x) = \lim_{x \to 100^+} (12x + k) = 12(100) + k = 1200 + k$.
3. Set the limits equal to each other: $2000 = 1200 + k$.
4. Solve for $k$: $k = 2000 - 1200 = 800$. Thus, $k$ must be 800 for the function to be continuous.
The profit function $P(x)$ for a certain product is known to be continuous for all $x \ge 0$, where $x$ is the number of units produced. What is the most significant business implication of this continuity?
Answer: The company can make small adjustments to production levels without causing sudden, drastic shocks to profit.
Continuity of a function, in a practical sense, means that small changes in the input result in small changes in the output. For a profit function $P(x)$, this means that if the production level $x$ is changed by a small amount, the profit $P(x)$ will also change by a small amount. This implies predictability and stability; there are no hidden 'jumps' or 'holes' where producing one more unit leads to a surprisingly different profit. Choice A is incorrect; continuity doesn't guarantee positivity. Choice C refers to the existence of a maximum, which is not guaranteed by continuity alone (differentiability is usually assumed). Choice D is incorrect; continuity does not imply a constant marginal cost.
A function $h(x)$ has a discontinuity at $x=c$. It is known that $\lim_{x \to c} h(x)$ exists and is a finite value, but $h(c)$ is undefined. What conclusion can be drawn about the discontinuity at $x=c$?
Answer: It is a removable discontinuity that can be eliminated by defining $h(c)$ to equal the limit.
The definition of a removable discontinuity at a point $x=c$ is that the limit, $\lim_{x \to c} h(x)$, exists and is finite, but either $h(c)$ is not defined or $h(c) \neq \lim_{x \to c} h(x)$. The question states that the limit exists but $h(c)$ is undefined. This fits the definition of a removable discontinuity perfectly. It is 'removable' because one could define $h(c)$ to be equal to the limit, making the function continuous at that point.
A company models its daily revenue, $R(t)$, as a function of time $t$ in days with $R(t) = \frac{|t-5|(t+2)}{t^2-3t-10}$. Analyze the discontinuities of this function.
Answer: A jump discontinuity at $t=5$ and a removable discontinuity at $t=-2$.
First, factor the denominator: $t^2-3t-10 = (t-5)(t+2)$. The function is $R(t) = \frac{|t-5|(t+2)}{(t-5)(t+2)}$. Potential discontinuities are at $t=5$ and $t=-2$.
1. Analyze at $t=-2$: The factor $(t+2)$ appears in both the numerator and denominator. $\lim_{t \to -2} \frac{|t-5|}{t-5} = \frac{|-7|}{-7} = \frac{7}{-7} = -1$. Since the limit exists, the discontinuity at $t=-2$ is removable.
2. Analyze at $t=5$: We need to check one-sided limits due to the absolute value $|t-5|$.
- Right-hand limit: $\lim_{t \to 5^+} \frac{(t-5)(t+2)}{(t-5)(t+2)} = \lim_{t \to 5^+} 1 = 1$.
- Left-hand limit: $\lim_{t \to 5^-} \frac{-(t-5)(t+2)}{(t-5)(t+2)} = \lim_{t \to 5^-} -1 = -1$.
Since the left and right-hand limits exist but are not equal ($-1 \neq 1$), the discontinuity at $t=5$ is a jump discontinuity.
A tiered pricing model for a software subscription is based on the number of users, $n$. The annual cost per user, $P(n)$, is given by $P(n) = 500 - 2\lfloor n/100 \rfloor$, where $\lfloor \cdot \rfloor$ is the floor function. What type of discontinuity does the total cost function, $C(n) = n \cdot P(n)$, have at $n=100$?
Answer: A jump discontinuity because the total cost abruptly changes at 100 users.
Let's analyze the total cost function $C(n) = n(500 - 2\lfloor n/100 \rfloor)$ around $n=100$. We need to check the one-sided limits.
1. Left-hand limit: As $n \to 100^-$ (e.g., $n=99.99$), $n/100$ is slightly less than 1, so $\lfloor n/100 \rfloor = 0$.
$\lim_{n \to 100^-} C(n) = \lim_{n \to 100^-} n(500 - 2(0)) = 100(500) = 50000$.
2. Right-hand limit: As $n \to 100^+$ (e.g., $n=100.01$), $n/100$ is slightly more than 1, so $\lfloor n/100 \rfloor = 1$. The value at $n=100$ also uses $\lfloor 100/100 \rfloor = 1$.
$\lim_{n \to 100^+} C(n) = \lim_{n \to 100^+} n(500 - 2(1)) = 100(498) = 49800$.
Since the left-hand limit ($50,000$) and the right-hand limit ($49,800$) are different, the function has a jump discontinuity at $n=100$. The total cost for 99 users is nearly $50,000, but the cost for 100 users is $49,800.
For the function $f(x) = \frac{x^2+ax-6}{x-2}$ to have a removable discontinuity at $x=2$, what must be the value of the limit $\lim_{x \to 2} f(x)$?
Answer: The limit must be 5.
When you encounter a rational function with a potential discontinuity, you're dealing with limits and continuity concepts. A removable discontinuity occurs when a function has a "hole" at a point that could theoretically be "filled in" to make the function continuous there.
For f(x)=x−2x2+ax−6 to have a removable discontinuity at x=2, the numerator must also equal zero when x=2. This creates the indeterminate form 00, which can potentially be resolved through factoring.
Setting the numerator equal to zero at x=2: 22+a(2)−6=0, which gives us 4+2a−6=0, so a=1.
With a=1, our function becomes f(x)=x−2x2+x−6. Factoring the numerator: x2+x−6=(x+3)(x−2).
Now we can simplify: f(x)=x−2(x+3)(x−2)=x+3 (for x=2).
Therefore, limx→2f(x)=2+3=5.
Looking at the wrong answers: B) 3 might result from incorrectly evaluating just the x+3 part at x=0. C) 1 could come from mishandling the algebra or using the wrong value of a. D) -1 might arise from sign errors in factoring or substitution.
Study tip: For removable discontinuities in rational functions, always check if both numerator and denominator are zero at the problem point. If so, factor and cancel the common terms to find the limit.
A company's marginal cost function has the form MC(x)=(x−4)2x2−16. Management wants to know if the discontinuity at x=4 can be 'repaired' by redefining the function at that point. What should an analyst conclude?
Answer: The discontinuity is non-removable because after factoring, (x−4) remains in the denominator with power 1
When analyzing discontinuities in rational functions, you need to determine whether the problematic point creates a "hole" that can be filled (removable) or an asymptote that cannot be fixed (non-removable). The key is examining what happens after factoring.
Let's factor the numerator: x2−16=(x+4)(x−4). So our function becomes:
MC(x)=(x−4)2(x+4)(x−4)=x−4x+4 (for x=4)
After canceling one factor of (x−4), we still have (x−4) in the denominator with power 1. This means as x approaches 4, the denominator approaches zero while the numerator approaches 4+4=8. This creates a vertical asymptote, making the discontinuity non-removable.
Answer A incorrectly suggests that any factoring and canceling makes a discontinuity removable. While we do eliminate one power of (x−4), what remains in the denominator determines removability.
Answer B makes a false generalization about rational functions. Limits don't automatically exist just because numerators approach zero—what matters is the behavior of both numerator and denominator.
Answer D focuses on degree comparison, which isn't the relevant criterion here. The key isn't comparing degrees but whether any factors remain in the denominator after canceling.
Study tip: For rational function discontinuities, always factor completely and cancel common terms. If nothing remains in the denominator at the problem point, it's removable. If factors remain, it's non-removable with a vertical asymptote.
A function modeling a company's market share percentage over time $t$ (in years) is given by $f(t) = \frac{t^2-t-12}{t^2-9}$. Which statement accurately describes the discontinuities of this function for $t>0$?
Answer: An infinite discontinuity at $t=3$ and no other discontinuities for $t>0$.
First, factor the numerator and the denominator of the function $f(t) = \frac{t^2-t-12}{t^2-9}$.
Numerator: $t^2-t-12 = (t-4)(t+3)$.
Denominator: $t^2-9 = (t-3)(t+3)$.
So, $f(t) = \frac{(t-4)(t+3)}{(t-3)(t+3)}$.
Potential discontinuities exist where the denominator is zero: $t=3$ and $t=-3$. Since the question specifies $t>0$, we only need to consider $t=3$.
At $t=-3$, the factor $(t+3)$ would cancel, creating a removable discontinuity, but this is outside the domain of interest.
At $t=3$, the denominator is zero, but the numerator is $(3-4)(3+3) = (-1)(6) = -6$. Since the limit approaches $\frac{-6}{0}$, this is an infinite discontinuity (a vertical asymptote). The point $t=4$ makes the numerator zero, but this is a root of the function, not a discontinuity. Therefore, the only discontinuity for $t>0$ is an infinite one at $t=3$.
The function f(x)=xtan(x) has a discontinuity at x=0. If we want to make this function continuous at x=0 by defining f(0)=k for some constant k, what value of k is required?
Answer: k=1 because limx→0xtan(x)=limx→0xsin(x)⋅cos(x)1=1⋅1=1
When you encounter a function that's undefined at a point but you need to make it continuous, you're dealing with a removable discontinuity. The key is finding the limit as x approaches that point.
Here, f(x)=xtan(x) is undefined at x=0 because both the numerator and denominator equal zero, creating the indeterminate form 00. To find the appropriate value of k, we need limx→0xtan(x).
Since tan(x)=cos(x)sin(x), we can rewrite this as:
limx→0xtan(x)=limx→0xcos(x)sin(x)=limx→0xsin(x)⋅cos(x)1
Using the fundamental limit limx→0xsin(x)=1 and the fact that cos(0)=1, we get 1⋅1=1. Therefore, k=1.
Answer A incorrectly suggests using L'Hôpital's rule would give 0, but L'Hôpital's rule would actually yield limx→01sec2(x)=1. Answer B is wrong because while tangent has discontinuities at x=2π+nπ, the discontinuity at x=0 is removable since the limit exists. Answer D incorrectly brings in the period of tangent, which is irrelevant to this local limit calculation.
Study tip: When you see 00 forms involving trig functions at x=0, look for ways to use the standard limits limx→0xsin(x)=1 and limx→0cos(x)=1.
A company's profit function is P(t)=⎩⎨⎧2t+110t2−2t+4if t<3if t=3if t>3 where t represents time in years. To properly classify the discontinuity at t=3, which analysis is required?
Answer: Evaluate limt→3−P(t)=7, limt→3+P(t)=7, and P(3)=10 to identify a removable discontinuity
To classify a discontinuity, we need all three values: limt→3−P(t)=2(3)+1=7, limt→3+P(t)=32−2(3)+4=7, and P(3)=10. Since both one-sided limits exist and are equal, but the function value differs, this is a removable discontinuity. Choice A is wrong because the function value matters for classification. Choice C is wrong because piecewise functions don't automatically have infinite discontinuities. Choice D is wrong because continuity of individual pieces doesn't guarantee continuity at boundary points.
