Precalculus Quiz: Constructing Inverse Trigonometric Functions

What this quiz covers

This quiz focuses on Constructing Inverse Trigonometric Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

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.

Question 1

1. Incorrect, because $\cos(x)$ is increasing on this interval, not decreasing
2. Correct, because $\cos(x)$ is strictly decreasing on $[\pi, 2\pi]$ (correct answer)
3. Incorrect, because the range is $[-1, 1]$ instead of $[1, -1]$
4. Correct, but only if we also restrict the range to $[0, 1]$

Explanation: The claim is correct. On the interval [π, 2π], cos(x) is strictly decreasing, going from cos(π) = -1 to cos(2π) = 1. Since the function is strictly monotonic (decreasing) on this interval, it is one-to-one and therefore has an inverse. Choice A incorrectly states that cos(x) is increasing on this interval. Choice C misunderstands that the range being [-1, 1] doesn't prevent inverse function construction. Choice D incorrectly suggests additional range restriction is needed when the monotonic property already ensures invertibility.

Question 2

1. The function must achieve its maximum and minimum values at the endpoints
2. The function must be either strictly increasing or strictly decreasing throughout (correct answer)
3. The function must have a horizontal tangent line at exactly one point
4. The function must be symmetric about the y-axis within the given domain

Explanation: For an inverse to exist, g(x) must be one-to-one on the given interval, which requires it to be either strictly increasing or strictly decreasing throughout the interval. Since g'(x) = 6cos(3x) and cos(3x) > 0 for x ∈ [-π/6, π/6] (as 3x ∈ [-π/2, π/2]), the function is strictly increasing. Choice A is incorrect because extrema location doesn't determine invertibility. Choice C is incorrect because having exactly one horizontal tangent would actually indicate the function changes from increasing to decreasing. Choice D is incorrect because symmetry would violate the one-to-one property.

Question 3

1. $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ because $\cos(x)$ increases from $0$ to $0$ on this interval
2. $\left[\frac{3\pi}{2}, \frac{5\pi}{2}\right]$ because $\cos(x)$ increases from $0$ to $0$ on this interval
3. $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ because $\cos(x)$ increases from $0$ to $0$ on this interval
4. $[\pi, 2\pi]$ because $\cos(x)$ increases from $-1$ to $1$ on this interval (correct answer)

Explanation: When creating an inverse function, you need the original function to be one-to-one (pass the horizontal line test) on your chosen domain. This means the function must be strictly monotonic - either strictly increasing or strictly decreasing throughout the interval. The standard restriction $[0, \pi]$ makes $\cos(x)$ strictly decreasing, but this question asks where cosine is strictly increasing instead. To find this, you need to analyze cosine's behavior over different intervals. On the interval $[\pi, 2\pi]$, cosine starts at $\cos(\pi) = -1$ and increases continuously to $\cos(2\pi) = 1$. Throughout this entire interval, the function has a positive derivative ($\cos'(x) = -\sin(x)$, and $\sin(x) < 0$ for $x \in (\pi, 2\pi)$), confirming it's strictly increasing. This makes choice D correct. Let's examine why the other options fail: Choice A claims $\cos(x)$ increases from 0 to 0 on $[\frac{\pi}{2}, \frac{3\pi}{2}]$, but cosine actually goes from 0 to -1 to 0, creating a decrease then increase - not strictly monotonic. Choice B has the same "0 to 0" error on $[\frac{3\pi}{2}, \frac{5\pi}{2}]$, where cosine goes from 0 to 1 to 0. Choice C incorrectly states cosine increases on $[-\frac{\pi}{2}, \frac{\pi}{2}]$, but cosine actually decreases on this interval (from 0 to 1 to 0, with the peak at $x = 0$). Study tip: When finding appropriate domains for inverse trig functions, sketch the graph and identify intervals where the function is strictly monotonic. Don't rely on endpoint values alone - check the behavior throughout the entire interval.

Question 4

1. Inverse functions require the original function to be bounded on the restricted domain
2. Inverse functions require the range to include only positive values
3. Inverse functions require the domain to be symmetric about the origin
4. Inverse functions require strict monotonicity, not approximate or partial monotonicity (correct answer)

Explanation: When creating inverse functions, you're essentially "undoing" a function by swapping its inputs and outputs. For this to work mathematically, the original function must pass a crucial test: it must be strictly monotonic (either always increasing or always decreasing) on its restricted domain. The correct answer is D because "mostly decreasing" isn't sufficient. On the interval $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$, $\sin(x)$ decreases from $\frac{\pi}{2}$ to $\pi$, but then increases from $\pi$ to $\frac{3\pi}{2}$. This creates a problem: multiple x-values produce the same y-value. For instance, both $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ give $\sin(x) = \frac{\sqrt{3}}{2}$. An inverse function couldn't determine which x-value to return, violating the definition of a function. Answer A is incorrect because boundedness isn't the issue—many valid inverse functions have unbounded domains. Answer B is wrong because inverse functions don't require positive ranges; the standard arcsine has range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, which includes negative values. Answer C is false because symmetry about the origin isn't required—the standard arcsine domain $[-1,1]$ is symmetric, but this isn't a general requirement. Remember: for inverse functions, strict monotonicity is non-negotiable. "Mostly" or "approximately" monotonic won't work—even one local maximum or minimum in your restricted domain will create the multiple-input problem that breaks the inverse relationship.

Question 5

1. It must be one-to-one (pass the horizontal line test). (correct answer)
2. It must be periodic.
3. It must be even.
4. It must have domain $[-1,1]$.

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. A function has an inverse only if it is one-to-one (each output corresponds to exactly one input, passing the horizontal line test), and since sine, cosine, and tangent are periodic functions that repeat values, we must restrict their domains to intervals where they are always increasing or always decreasing. The restriction to [-π/2, π/2] is chosen because on this interval, sine is always increasing (never decreases): as x increases from -π/2 to π/2, sin(x) increases from -1 to 1 without ever going back down, ensuring one-to-one correspondence. Choice A is correct because it properly identifies the one-to-one requirement for invertibility. Choice B incorrectly claims that periodicity makes the function invertible, when in fact periodicity (repeating values) is exactly why we need to restrict the domain to create a one-to-one function. The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.

Question 6

1. $\pi/3$ (correct answer)
2. $2\pi/3$
3. $\pi/6$
4. $5\pi/6$

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). To evaluate arcsin([value]), we ask: what angle in [-π/2, π/2] has sine equal to [value]? Using special triangle values, we know sin(π/[6 or 4 or 3]) = [value], and since π/[6 or 4 or 3] is in the range of arcsin, we have arcsin([value]) = π/[6 or 4 or 3]. Choice A is correct because it correctly evaluates using restricted range. Choice B gives a value outside the range of arcsin, stating arcsin(1/2) = [5π/6 or other], but arcsin must return an angle in [-π/2, π/2], and [wrong value] is outside this interval. To evaluate inverse trig functions at special values, use your knowledge of special angles: arcsin(1/2) = π/6, arcsin(√2/2) = π/4, arcsin(√3/2) = π/3, and remember the answer must lie in the appropriate restricted range.

Question 7

1. Because $\sin(x)$ must be one-to-one (pass the horizontal line test) on the restricted domain to have an inverse function. (correct answer)
2. Because $\sin(x)$ is undefined outside $[-\pi/2,\pi/2]$.
3. Because restricting the domain makes the range of $\sin(x)$ larger than $[-1,1]$.
4. Because periodic functions automatically have inverses on their full domains.

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. A function has an inverse only if it is one-to-one (each output corresponds to exactly one input, passing the horizontal line test), and since sine, cosine, and tangent are periodic functions that repeat values, we must restrict their domains to intervals where they are always increasing or always decreasing. Without restriction, sine takes the same value at multiple angles (for example, sin(π/6) = sin(5π/6) = 1/2), so given an output like 1/2, we couldn't determine a unique input—the function wouldn't pass the horizontal line test. By restricting to [-π/2, π/2], each output value occurs at exactly one input, making the function invertible. Choice A is correct because it properly identifies the one-to-one requirement. Choice D incorrectly claims that periodicity makes the function invertible, when in fact periodicity (repeating values) is exactly why we need to restrict the domain to create a one-to-one function. The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing. Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

Question 8

1. No inverse exists because the function is not continuous at $x = \frac{\pi}{2}$
2. No inverse exists because both pieces achieve the same output values (correct answer)
3. An inverse exists because each piece is strictly monotonic on its domain
4. An inverse exists because the function is defined on a closed interval

Explanation: The function has no inverse because it fails the horizontal line test. Since both sin(x) on [0, π/2] and cos(x) on (π/2, π] can achieve the same output values (for example, sin(π/6) = 1/2 and cos(π/3) = 1/2), the function is not one-to-one across its entire domain. Choice A is incorrect because continuity isn't required for inverse existence, only the one-to-one property. Choice C is incorrect because individual monotonicity of pieces doesn't guarantee global one-to-one property. Choice D is incorrect because being defined on a closed interval doesn't ensure invertibility.

Question 9

1. The function would not be one-to-one because $\sin(0) = \sin(\pi) = 0$ (correct answer)
2. The function would not be continuous on the given interval
3. The range would not include all values between $-1$ and $1$
4. The function would not be differentiable at the endpoints

Explanation: For an inverse function to exist, the original function must be one-to-one (injective) on its restricted domain. On [0, π], sin(x) takes the value 0 at both x = 0 and x = π, violating the one-to-one property. The standard restriction [-π/2, π/2] ensures sin(x) is strictly increasing and therefore one-to-one. Choice B is incorrect because sin(x) is continuous everywhere. Choice C is incorrect because sin(x) does achieve its full range [-1, 1] on [0, π]. Choice D is incorrect because differentiability at endpoints is not required for inverse function existence.

Question 10

1. Because restricting the domain makes $\sin(x)$ one-to-one (it passes the horizontal line test) on $[-\pi/2,\pi/2]$. (correct answer)
2. Because $\sin(x)$ is periodic, it is automatically one-to-one on all real numbers.
3. Because the range of $\sin(x)$ becomes $[-\pi/2,\pi/2]$ after restriction.
4. Because $\sin(x)$ still fails the horizontal line test on $[-\pi/2,\pi/2]$ unless we also restrict the range.

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. A function has an inverse only if it is one-to-one (each output corresponds to exactly one input, passing the horizontal line test), and since sine, cosine, and tangent are periodic functions that repeat values, we must restrict their domains to intervals where they are always increasing or always decreasing. Without restriction, sine takes the same value at multiple angles (for example, sin(π/6) = sin(5π/6) = 1/2), so given an output like 1/2, we couldn't determine a unique input—the function wouldn't pass the horizontal line test. By restricting to [-π/2, π/2], each output value occurs at exactly one input, making the function invertible. Choice A is correct because it properly identifies the one-to-one requirement. Choice B incorrectly claims that periodicity makes the function invertible, when in fact periodicity (repeating values) is exactly why we need to restrict the domain to create a one-to-one function. The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.

Question 11

1. Check if $p'(x) = \cos(x + \frac{\pi}{4})$ maintains the same sign throughout the interval (correct answer)
2. Verify that the function achieves its maximum value at exactly one endpoint
3. Confirm that the horizontal line $y = 0$ intersects the graph exactly once
4. Show that the function is symmetric about the line $x = -\frac{\pi}{4}$

Explanation: The most direct way to verify that a function is one-to-one (and thus has an inverse) is to check that its derivative maintains the same sign throughout the interval, ensuring strict monotonicity. For p(x) = sin(x + π/4), we have p'(x) = cos(x + π/4). On [-3π/4, π/4], we have x + π/4 ∈ [-π/2, π/2], where cos is positive except at the endpoints, so p(x) is strictly increasing. Choice B is insufficient because endpoint behavior doesn't guarantee monotonicity. Choice C tests only one specific value, not the general one-to-one property. Choice D describes symmetry, which would actually violate the one-to-one property.

Question 12

1. $[-\pi/2,\pi/2]$
2. $[-1,1]$ (correct answer)
3. $[0,\pi]$
4. $(-\infty,\infty)$

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). The domain of arcsin is [-1, 1] because these are the possible output values of the sine function (sin(x) always lies between -1 and 1). Choice B is correct because it correctly gives the domain of the inverse as [-1,1]. Choice A confuses the domain and range of arcsin, stating the domain is [-π/2, π/2] when actually that's the range—the domain of arcsin is [-1, 1] (the possible sine values). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

Question 13

1. The function must be continuous, but $\sin(x)$ has a discontinuity at $x = \pi$
2. The function must have a finite range, but $\sin(x)$ approaches infinity
3. The function must be strictly monotonic, but $\sin(x)$ both decreases and increases (correct answer)
4. The function must be differentiable, but $\sin(x)$ has a corner at $x = \frac{3\pi}{2}$

Explanation: When determining if a function has an inverse, you need to check if it's one-to-one (bijective). This means each output value corresponds to exactly one input value, which requires the function to be strictly monotonic—either always increasing or always decreasing throughout its domain. Looking at $\sin(x)$ on $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$, the function starts at 1, decreases to -1 at $x = \pi$, then increases back to 0. This creates a problem: the function takes on the same y-values (like $y = 0$) at multiple x-values within this interval. For instance, $\sin\left(\frac{\pi}{2} + \epsilon\right) = \sin\left(\frac{3\pi}{2} - \epsilon\right)$ for small positive $\epsilon$. This violates the one-to-one requirement because an inverse function wouldn't know which x-value to return for these repeated y-values. Answer C correctly identifies that the function must be strictly monotonic, which $\sin(x)$ fails to be on this interval since it both decreases and increases. Answer A is wrong because $\sin(x)$ is continuous everywhere, including at $x = \pi$. Answer B is incorrect since $\sin(x)$ has a finite range $[-1,1]$ and never approaches infinity. Answer D is false because $\sin(x)$ is differentiable everywhere and has no corners—its derivative exists at every point. Remember: for inverse functions, look for strict monotonicity first. The standard domain restriction for $\arcsin$ uses $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ precisely because $\sin(x)$ is strictly increasing there.

Question 14

1. $\tan(x)$ is always increasing on $(-\pi/2,\pi/2)$, so it is one-to-one there. (correct answer)
2. $\tan(x)$ is periodic, so it is one-to-one on all real numbers.
3. $\tan(x)$ has range $(-\pi/2,\pi/2)$ on $(-\pi/2,\pi/2)$, so it is invertible.
4. $\tan(x)$ is always decreasing on $(-\pi/2,\pi/2)$, so it is one-to-one there.

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse tangent function arctan(x) is constructed by restricting tan(x) to (-π/2, π/2) where tangent is always increasing, and since tangent's range on this interval is all real numbers, arctan has domain (-∞, ∞) and range (-π/2, π/2). The restriction to (-π/2, π/2) is chosen because on this interval, tangent is always increasing (never decreases): as x increases from -π/2 to π/2, tan(x) increases from -∞ to ∞ without ever going back down, ensuring one-to-one correspondence. Choice A is correct because it properly identifies the always increasing property that makes it one-to-one. Choice D claims that tangent is always decreasing on the restricted interval, but actually it's always increasing on (-π/2, π/2), which is what makes it one-to-one. The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.

Question 15

1. The function $\tan(x)$ is not actually increasing on the interval $[\pi, 2\pi]$
2. The union of two intervals cannot serve as a domain for inverse functions
3. The function would not be one-to-one across the entire proposed domain (correct answer)
4. The discontinuity at $x = \pi$ prevents the inverse from being well-defined

Explanation: While tan(x) is indeed increasing on each subinterval [0, π) and (π, 2π], the function is not one-to-one across the entire union because tan(x) repeats its values. For example, tan(π/4) = 1 and tan(5π/4) = 1, so the value 1 would correspond to two different inputs. Choice A is incorrect because tan(x) is increasing on [π, 2π]. Choice B is incorrect because unions can serve as domains if the overall function is one-to-one. Choice D is incorrect because the discontinuity itself doesn't prevent inverse construction if the one-to-one property is maintained.

Question 16

1. $[-1,1]$
2. $[0,\pi]$
3. $[-\pi/2,\pi/2]$ (correct answer)
4. $(-\infty,\infty)$

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). The range of arcsin is [-π/2, π/2] because these are the angles we allow in the restricted domain of sine that we're inverting. Choice C is correct because it correctly gives the range of the inverse as [-π/2, π/2]. Choice A confuses the domain and range of arcsin, stating the range is [-1, 1] when actually that's the domain—the range of arcsin is [-π/2, π/2] (the restricted domain of sin). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

Question 17

1. $[-\pi/2,\pi/2]$
2. $[-1,1]$ (correct answer)
3. $( -\pi,\pi )$
4. $(-\infty,\infty)$

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). The domain of arcsin is [-1, 1] because these are the possible output values of the sine function (sin(x) always lies between -1 and 1). Choice B is correct because it correctly states the domain of the inverse as [-1, 1]. Choice A confuses the domain and range of arcsin, stating the domain is [-π/2, π/2] when actually that's the range—the domain of arcsin is [-1, 1] (the possible sine values). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

Question 18

1. $x\in[-1,1]$
2. $x\in[0,\pi]$
3. $x\in(-\infty,\infty)$
4. $x\in[-\pi/2,\pi/2]$ (correct answer)

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse sine function arcsin(x) is constructed by restricting sin(x) to the domain [-π/2, π/2] where sine is always increasing, making it one-to-one and therefore invertible; the resulting inverse function arcsin has domain [-1, 1] (the range of sin) and range [-π/2, π/2] (the restricted domain of sin). For sin(arcsin(x)) where x ∈ [-1, 1], arcsin(x) gives an angle in [-π/2, π/2], and taking sine of that angle returns the original value x. However, arcsin(sin(x)) = x only when x is already in [-π/2, π/2]; for other values like x = 3π/4, arcsin(sin(3π/4)) gives π/4 (the angle in the restricted range with the same sine value). Choice D is correct because it correctly states the interval where the composition equals x as [-π/2, π/2]. Choice A confuses the domain of arcsin (which is [-1, 1] for the input to arcsin) with the values of x where arcsin(sin(x)) = x holds. For composition: f(f⁻¹(x)) = x always works (for x in the domain of f⁻¹), but f⁻¹(f(x)) = x only when x is in the restricted domain—outside the restricted interval, the inverse function returns an equivalent angle in the restricted range.

Question 19

1. $[-1,1]$
2. $[-\pi/2,\pi/2]$
3. $[0,\pi]$ (correct answer)
4. $(-\infty,\infty)$

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse cosine function arccos(x) uses the restriction of cos(x) to [0, π] where cosine is always decreasing (one-to-one), giving arccos a domain of [-1, 1] and range of [0, π]. The range of arccos is [0, π] because these are the angles we allow in the restricted domain of cosine that we're inverting. Choice C is correct because it correctly gives the range of the inverse as [0, π]. Choice B confuses the domain and range of arccos, stating the range is [-π/2, π/2] when actually that's the range for arcsin—the range of arccos is [0, π] (the restricted domain of cos). Key to inverse trig functions: remember that arcsin has domain [-1, 1] and range [-π/2, π/2], arccos has domain [-1, 1] and range [0, π], and arctan has domain all reals and range (-π/2, π/2)—the domain is always the possible trig output values, and the range is the restricted input interval.

Question 20

1. $[-\pi/2,\pi/2]$
2. [0,\pi] (correct answer)
3. $( -\pi,\pi )$
4. [0,2\pi]

Explanation: This question tests understanding of how inverse trigonometric functions are constructed by restricting domains. The inverse cosine function $\arccos(x)$ uses the restriction of $\cos(x)$ to $[0, \pi]$ where cosine is always decreasing (one-to-one), giving $\arccos$ a domain of $[-1, 1]$ and range of $[0, \pi]$. Without restriction, cosine takes the same value at multiple angles (for example, $\cos(\pi/3) = \cos(-\pi/3) = 1/2$), so given an output like 1/2, we couldn't determine a unique input—the function wouldn't pass the horizontal line test. By restricting to $[0, \pi]$, each output value occurs at exactly one input, making the function invertible. Choice B is correct because it correctly states the restricted domain for $\arccos$ as $[0, \pi]$. Choice A confuses the restriction for arcsin (which is $[-\pi/2, \pi/2]$) with the restriction for $\arccos$ (which is $[0, \pi]$). The restriction is necessary because a function can only have an inverse if it's one-to-one (passes the horizontal line test), and sine, cosine, and tangent are periodic (repeating), so they're only one-to-one on restricted intervals where they're always increasing or always decreasing.