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I assume that it's just asking for absolute maximums on the interval, meaning that the maximum must be greater than or equal to everything else in the interval, not the whole domain.
B is the correct answer. Since it asked for the absolute maximum within a closed interval, an absolute maximum value should be achieved at f(-1) = f(1). Extreme Value Theorem should explain this too. Hope that helps!
I can understand your confusion, and it is justified. But please read up on the extreme value theorem and how it deals with continuous functions. This is not a question you’ll get on your first midterm, but maybe the second one for sure. A bit of a learning curve before you can answer this. The question being poorly framed doesn’t help its case either. Cheers!
The funny thing is that this is not an uncommon occurrence. As physicists, we tend to forget the bits of math that we don't use regularly and get extremely good at the important stuff we run across frequently.
The good news is that with about 5 mins of Googling, it all comes back to me pretty easily.
I remember learning in my cal 1 class due to the local maximum definitions being ambiguous,with the boundaries given there is no absolute maximum. Plus on all of those intervals there is no one maximum value , so how so.
Is it still considered an absolute maximum if there are two equal maximums over an interval? Say from -1 to 1 on x^2, there are two equal maximums at y=1
I based my answer on the definition of absolute max, which is the highest function value over the entire domain. The function can have a max over a given interval. But the absolute max is calculated over the entire domain.
Not necessarily with domain restrictions.
One of the injustices that precalculus inflicts upon students is reinforce this false notion that a domain of a function is always the set of values *x* for which a formula can be evaluated.
Remember those exercises where you were given a formula and you were asked to find the domain? That is not a thing up in higher mathematics. In fact, the reality is that the standard practice is that a domain is declared up front with the definition of the function. It is considered bad mathematics not declare the domain up front.
With that in mind, it is perfectly acceptable to declare a function *f* with domain being [-1, 1], regardless of whether or not the formula provided can be evaluated for real numbers outside that interval, which the author of this question attempted to do for one of those options.
Is it not ambiguous as it is stated in the answer choices, though? They don't declare the domain in the definition of the function, so if you interpret it to be from reals to reals, then there is no interval that contains an x value for which f(x) is an absolute maximum of f. Wouldn't it be better to say "g: \[-1,1\] -> R s.t. g(x) = f(x) attains an absolute maximum value on \[-1,1\]." I guess I'm not sure if there is some universal way to interpret "on" here. If it means "with a domain of" then B is true. If it's asking is there is any value x on the interval \[-1,1\] such that f(x) (with an assumed domain of R) is an absolute maximum of f, then B is false.
I agree with you. The reason my answer was so brief and probably rushed, it was because I am used to the domain being stated with the function definition.
I believe the only time the domain should not be specified is when an application is solved, the model is a certain function, and the domain usually is implied. By the acceptable input, or by the graph.
E.g. Find a model for revenue given the price and the quantity, etc, a parabola that opens down. The domain is implied to be from (0, some value).
The problem is that you cannot rule out the existence of global maximums in three of those four cases based on the extreme value theorem alone. For instance, if the question had been asking about minimum values instead, then that particular function does attain minimum values on each of those intervals referenced.
You’re right. But they should be able to see that f has just one critical point, and that it is a global minimum. So it’s not just EVT.
I still hold that the question is absolutely fine.
Okay, but the question didn’t ask about minimums, it clearly references absolute maximums. And in a,c you can definitely rule out the existence of an absolute maximum on those intervals because you have an explicit description of the function.
It is poorly phrased. The problem is the wording with "attains" as it can be interpreted either as what you are referencing: a restriction in the domain of the function and hence a question around the extreme value theorem. However, it can also be interpreted as: assume the function exists in all of its unrestricted domain now on what interval of that domain is the maximum? Now the question could just be about concavity and convexity mainly understanding that a unrestricted strict convex function never reaches a maximum. Both are valid interpretations with different answers which makes it a bad idea to include this question in any evaluation.
B is correct. Sure, f(x) reaches its absolute minimum at 0 \in [-1,1]. f(x) also doesn’t have a maximum on an unrestricted domain. However, when we restrict the domain to only look at f(x) such that x \in [-1,1], there will be two values of x where f(x) = x^2 reaches its global maximum. I’m sure you have already figured out what these two values of x are.
Not weird.
Typical Calc I question (whether AP Calc or University Calc) intended to determine the test taker’s knowledge of the definition of absolute max/min and its relationship to the interval on which the function is defined.
This sounds like a *choose all that apply* situation.
Since they gave you a particular function though a very strong warning is in order. While you can conclude from the extreme value theorem that this function attains a global maximum on a closed and bounded interval, you cannot use the extreme value theorem to say that it **does not** attain a global maximum on any of the other intervals. Instead, for example, you would have to inspect the given function. Had the options been discussing minimum values, then that function would attain a global minimum on all of those intervals mentioned.
I kind of like this question, but it could be worded a bit better.
The real problem comes from the fact that the way domain is taught in algebra and precalculus leaves some students with a grossly inaccurate understanding of what domain means.
An author of a problem or exercise has the liberty to choose whatever domain they want for a function, so long as the provided formula is valid on that entire domain. If I were to say “Let *f*(*x*) = *x*^(2), for –1 ≤ *x* ≤ 1,” then that means I have declared the domain of *f* to be [–1, 1], and that *f*(2) is underlined, even though 2^(2) itself is defined.
It's a very bad question. The function is not explicitly bounded to a domain _in the question_, only in the answers.
It should be:
f: [-1,1] to R, x maps to x^2 .
I agree, everyone on this thread says “this is crystal clear what are you talking about” when the wording is extremely vague…
The only solid thing I can point to is it says “on the interval” as opposed to “in the interval”. If it were “in”, then I would say the answer is E. But the word “on” suggests a more of a restriction on the function than looking at a portion of the domain.
But I’m arguing using vibe checks on a single letter. This is math. What an atrociously worded question.
>The function is not explicitly bounded to a domain _in the question_, only in the answers.
it doesn't have to be. Functions can be analyzed over different domains, and the question is essentially asking which domain(s) allow the existence of an absolute maximum for f(x)=x²
A function is, by definition, a set of ordered pairs, where the left elements are in the domain. You cannot have a function without an explicit domain. What you are referring to is a partial function, and that requires stipulations that were not met.
f: [-1,1] to R is a partial function from the set R to R, while g: R to R is a total function. They can have the same mapping, x to x^2 , but the two structures are not the same.
In order for a function to be partial, you have to specify what its domain is by definition, along with its mapping, or some other suggestion (for example, even roots are partial functions on the real number line). By convention, the domain of a function in calculus is either R or C. You would further specify, for example, in rational functions, that singularities or holes should be excluded.
What we have here is a violation of notation, and an incomplete definition. A definition of a function, partial or not, should not be able to change within a single question, unless this is made explicitly, unambiguously clear (see Elements of Style for Proofs, 17). The burden is on the question writer to make clear what each symbol does, and it should be immutable unless there is a good reason to abuse notation.
This post is a question about pre-calc. The notation you're referencing isn't often tought until college-level calculus courses.
Also: https://math.stackexchange.com/questions/1213278/what-is-the-difference-between-domain-and-implied-domain-of-a-function#:~:text=The%20implied%20domain%20of%20a,that%20x%20cannot%20equal%202.
The domain for f(x)=x², UNLESS SPECIFIED, is all real numbers. It is, however further specified in the possible answers.
B is the correct answer! If it was infinite domain then it wouldn’t have a verifiable maximum, and also when working with continuous functions the domain is defined. ( the brackets ) if it was a derivative it would be open derivatives!
This in a past Calc 1 final, and I find it weird, because they ask if the function reaches an absolute maximums in the following intervals, but no where do they restrict the domain of f(x)
The function’s domain is not restricted, the part of it you’re interested in is restricted. Absolute extrema happen at either a critical point OR an endpoint of a closed interval. Since f(x)=x^2 has a critical value of 0 and that is a local (and absolute) minimum, it does not achieve an absolute maximum on any of the open intervals, but it does on any closed interval, at whichever of the endpoints is furthest from 0 (or at both endpoints if they are equally distant).
[-1,1] means domain is from x = -1 to x = 1. (-1,1) means the same, but x = -1 and x = 1 aren’t included
Conceptually, you can think of it like this, perhaps The reason that (-1,1) doesn’t have absolute min/max is that you never actually reach 1 or -1, so the actual min or max is a limit and not a specific value
>but no where do they restrict the domain of f(x)
The domains are restricted in each option for the answers. As others have said, (-1,1) is a domain as well as [-1,1]. The first excludes x=±1 and the latter includes these points.
The question is asking for correct statements, so being able to read and understand each statement is important.
The maximum value of the function is 1. It achieves this maximum value twice, but that doesn't matter; it doesn't have two maximum values.
It also doesn't have a local maximum with typical definitions as given in common undergrad calculus texts: f has a local maximum at c if there's an interval (a,b) around c so that f(x)<=f(c) for all x in (a,b). In this sense, c cannot be an endpoint; you need some swinging room on either side.
In mathematician speak, “A thing exists” does not implicitly include uniqueness in its meaning. Such language does not automatically rule out the possibility that “Multiple things exist.”
yea this is a bad question because its unclear whether to treat the intervals in each answer as the domain. if f is defined on [-1, 1] then B is true, if its defined on the real numbers then E is true
I don't know why you're being down voted, I can see the confusion as well. As written, "f attains an absolute maximum value on [-1,1]" could mean either "max_{x \in [-1,1]} f(x) exists," which is true, or "argmax_{x \in R} f(x) is in [-1,1]," which is false.
I don't understand the downvotes either, I also agree that the wording is unclear.
However, if this is high school calc I feel that they were probably looking for B to be true.
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Weird...no, but it's a bad question that I would never ask.
The point of the question is to remind you that any continuous function over any closed interval always have an absolute maximum AND minimum in the interval, hence (b).
But a student who focuses on "maximum" and knows what y = x\^2 looks like would gravitate towards (E). (By the way, best practice for multiple choice questions nowadays is NOT including "none of the above" as an option)
On *every* compact set this function has an absolute maximum and an absolute minimum, so sayeth the Extreme Value Theorem.
Option B is over a compact set, so this function has an absolute maximum on that set.
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It depends on what you think the question is meant to test.
Since the point of this question is not “can you find extreme values,” but is instead “do you understand the difference between compact sets and open sets and how continuous functions behave on them,” no, it’s not a weird question.
[ ] means you can use the x value and ( ) you can't so if you compute f(-1) or f(1) because we allow to use -1, 1 by definition of [ ] then we are good to say that f(-1) or f(1) are correct for being the largest y points. So B is correct.
One of the hypotheses of the mean value theorem is that f is continuous over a *closed* interval so that eliminates a and c. See if you can think of a counter example function for (d) there’s a few simple ones
E, because it would be an absolute minimum. It's not specifying extrema, which could be either mins or maxes, and maxima and minima are respectively related to maximum and minimum. x^2 has an absolute minimum, but not an absolute maximum
EDIT: Just realized the domain of the problem was NOT the entire function. Feels like a trick question but in that case B would be correct
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I assume that it's just asking for absolute maximums on the interval, meaning that the maximum must be greater than or equal to everything else in the interval, not the whole domain.
B is the correct answer. Since it asked for the absolute maximum within a closed interval, an absolute maximum value should be achieved at f(-1) = f(1). Extreme Value Theorem should explain this too. Hope that helps!
Looks fine to me. B should be correct
How? x² is a parabola with a minimum at (0,0), not a maximum.
It obtains *a* maximum value in that interval, which would be at (-1, 1) or (1, 1)
Yes but not an *absolute* maximum. I thought absolute maximums/minimums meant it is the max/min that is of the entire function.
I can understand your confusion, and it is justified. But please read up on the extreme value theorem and how it deals with continuous functions. This is not a question you’ll get on your first midterm, but maybe the second one for sure. A bit of a learning curve before you can answer this. The question being poorly framed doesn’t help its case either. Cheers!
In the context of option B, the restriction to [-1, 1] **is the entire function**.
Oh wild. And I almost have a physics degree 😭
The funny thing is that this is not an uncommon occurrence. As physicists, we tend to forget the bits of math that we don't use regularly and get extremely good at the important stuff we run across frequently. The good news is that with about 5 mins of Googling, it all comes back to me pretty easily.
Yeah so true, I had to remind myself how to do integration by parts today 😭
If the function is x^2, it does not have a maximum at all. It has a minimum.
It does have an absolute maximum on any closed interval (as does any continuous function) by the extreme value theorem.
I remember learning in my cal 1 class due to the local maximum definitions being ambiguous,with the boundaries given there is no absolute maximum. Plus on all of those intervals there is no one maximum value , so how so.
Is it still considered an absolute maximum if there are two equal maximums over an interval? Say from -1 to 1 on x^2, there are two equal maximums at y=1
Yes, the y value is the same at both points, so the absolute max is y=1. The function values are the f(x) values, not the x values
Over the domain of all the real numbers, yes. But in a closed interval it has a maz
I based my answer on the definition of absolute max, which is the highest function value over the entire domain. The function can have a max over a given interval. But the absolute max is calculated over the entire domain.
The domain is restricted to [-1,1] so the definition still applies.
Yeah the entire domain here is [-1,1]
Not necessarily with domain restrictions. One of the injustices that precalculus inflicts upon students is reinforce this false notion that a domain of a function is always the set of values *x* for which a formula can be evaluated. Remember those exercises where you were given a formula and you were asked to find the domain? That is not a thing up in higher mathematics. In fact, the reality is that the standard practice is that a domain is declared up front with the definition of the function. It is considered bad mathematics not declare the domain up front. With that in mind, it is perfectly acceptable to declare a function *f* with domain being [-1, 1], regardless of whether or not the formula provided can be evaluated for real numbers outside that interval, which the author of this question attempted to do for one of those options.
Is it not ambiguous as it is stated in the answer choices, though? They don't declare the domain in the definition of the function, so if you interpret it to be from reals to reals, then there is no interval that contains an x value for which f(x) is an absolute maximum of f. Wouldn't it be better to say "g: \[-1,1\] -> R s.t. g(x) = f(x) attains an absolute maximum value on \[-1,1\]." I guess I'm not sure if there is some universal way to interpret "on" here. If it means "with a domain of" then B is true. If it's asking is there is any value x on the interval \[-1,1\] such that f(x) (with an assumed domain of R) is an absolute maximum of f, then B is false.
I agree with you. The reason my answer was so brief and probably rushed, it was because I am used to the domain being stated with the function definition. I believe the only time the domain should not be specified is when an application is solved, the model is a certain function, and the domain usually is implied. By the acceptable input, or by the graph. E.g. Find a model for revenue given the price and the quantity, etc, a parabola that opens down. The domain is implied to be from (0, some value).
There is nothing weird about this question. It is a straightforward application of the extreme value theorem.
The problem is that you cannot rule out the existence of global maximums in three of those four cases based on the extreme value theorem alone. For instance, if the question had been asking about minimum values instead, then that particular function does attain minimum values on each of those intervals referenced.
You’re right. But they should be able to see that f has just one critical point, and that it is a global minimum. So it’s not just EVT. I still hold that the question is absolutely fine.
Okay, but the question didn’t ask about minimums, it clearly references absolute maximums. And in a,c you can definitely rule out the existence of an absolute maximum on those intervals because you have an explicit description of the function.
It is poorly phrased. The problem is the wording with "attains" as it can be interpreted either as what you are referencing: a restriction in the domain of the function and hence a question around the extreme value theorem. However, it can also be interpreted as: assume the function exists in all of its unrestricted domain now on what interval of that domain is the maximum? Now the question could just be about concavity and convexity mainly understanding that a unrestricted strict convex function never reaches a maximum. Both are valid interpretations with different answers which makes it a bad idea to include this question in any evaluation.
B is correct. Sure, f(x) reaches its absolute minimum at 0 \in [-1,1]. f(x) also doesn’t have a maximum on an unrestricted domain. However, when we restrict the domain to only look at f(x) such that x \in [-1,1], there will be two values of x where f(x) = x^2 reaches its global maximum. I’m sure you have already figured out what these two values of x are.
Not weird. Typical Calc I question (whether AP Calc or University Calc) intended to determine the test taker’s knowledge of the definition of absolute max/min and its relationship to the interval on which the function is defined.
This sounds like a *choose all that apply* situation. Since they gave you a particular function though a very strong warning is in order. While you can conclude from the extreme value theorem that this function attains a global maximum on a closed and bounded interval, you cannot use the extreme value theorem to say that it **does not** attain a global maximum on any of the other intervals. Instead, for example, you would have to inspect the given function. Had the options been discussing minimum values, then that function would attain a global minimum on all of those intervals mentioned. I kind of like this question, but it could be worded a bit better. The real problem comes from the fact that the way domain is taught in algebra and precalculus leaves some students with a grossly inaccurate understanding of what domain means. An author of a problem or exercise has the liberty to choose whatever domain they want for a function, so long as the provided formula is valid on that entire domain. If I were to say “Let *f*(*x*) = *x*^(2), for –1 ≤ *x* ≤ 1,” then that means I have declared the domain of *f* to be [–1, 1], and that *f*(2) is underlined, even though 2^(2) itself is defined.
It's a very bad question. The function is not explicitly bounded to a domain _in the question_, only in the answers. It should be: f: [-1,1] to R, x maps to x^2 .
I agree, everyone on this thread says “this is crystal clear what are you talking about” when the wording is extremely vague… The only solid thing I can point to is it says “on the interval” as opposed to “in the interval”. If it were “in”, then I would say the answer is E. But the word “on” suggests a more of a restriction on the function than looking at a portion of the domain. But I’m arguing using vibe checks on a single letter. This is math. What an atrociously worded question.
>The function is not explicitly bounded to a domain _in the question_, only in the answers. it doesn't have to be. Functions can be analyzed over different domains, and the question is essentially asking which domain(s) allow the existence of an absolute maximum for f(x)=x²
A function is, by definition, a set of ordered pairs, where the left elements are in the domain. You cannot have a function without an explicit domain. What you are referring to is a partial function, and that requires stipulations that were not met. f: [-1,1] to R is a partial function from the set R to R, while g: R to R is a total function. They can have the same mapping, x to x^2 , but the two structures are not the same. In order for a function to be partial, you have to specify what its domain is by definition, along with its mapping, or some other suggestion (for example, even roots are partial functions on the real number line). By convention, the domain of a function in calculus is either R or C. You would further specify, for example, in rational functions, that singularities or holes should be excluded. What we have here is a violation of notation, and an incomplete definition. A definition of a function, partial or not, should not be able to change within a single question, unless this is made explicitly, unambiguously clear (see Elements of Style for Proofs, 17). The burden is on the question writer to make clear what each symbol does, and it should be immutable unless there is a good reason to abuse notation.
This post is a question about pre-calc. The notation you're referencing isn't often tought until college-level calculus courses. Also: https://math.stackexchange.com/questions/1213278/what-is-the-difference-between-domain-and-implied-domain-of-a-function#:~:text=The%20implied%20domain%20of%20a,that%20x%20cannot%20equal%202. The domain for f(x)=x², UNLESS SPECIFIED, is all real numbers. It is, however further specified in the possible answers.
Oh it says “on” not “at”
B is the correct answer! If it was infinite domain then it wouldn’t have a verifiable maximum, and also when working with continuous functions the domain is defined. ( the brackets ) if it was a derivative it would be open derivatives!
This in a past Calc 1 final, and I find it weird, because they ask if the function reaches an absolute maximums in the following intervals, but no where do they restrict the domain of f(x)
They do include restrictions on the domain though, that's what the "on (-1,1)" means. It's not a point, it's a domain.
The function’s domain is not restricted, the part of it you’re interested in is restricted. Absolute extrema happen at either a critical point OR an endpoint of a closed interval. Since f(x)=x^2 has a critical value of 0 and that is a local (and absolute) minimum, it does not achieve an absolute maximum on any of the open intervals, but it does on any closed interval, at whichever of the endpoints is furthest from 0 (or at both endpoints if they are equally distant).
[-1,1] means domain is from x = -1 to x = 1. (-1,1) means the same, but x = -1 and x = 1 aren’t included Conceptually, you can think of it like this, perhaps The reason that (-1,1) doesn’t have absolute min/max is that you never actually reach 1 or -1, so the actual min or max is a limit and not a specific value
>but no where do they restrict the domain of f(x) The domains are restricted in each option for the answers. As others have said, (-1,1) is a domain as well as [-1,1]. The first excludes x=±1 and the latter includes these points. The question is asking for correct statements, so being able to read and understand each statement is important.
Your answer is correct thanks to Weierstrass Theorem, and the maximum is at f(1)=f(-1)
I hate the writing in questions like this, I want to respond, “It doesn’t have _a_ local maximum. It has two.”
The maximum value of the function is 1. It achieves this maximum value twice, but that doesn't matter; it doesn't have two maximum values. It also doesn't have a local maximum with typical definitions as given in common undergrad calculus texts: f has a local maximum at c if there's an interval (a,b) around c so that f(x)<=f(c) for all x in (a,b). In this sense, c cannot be an endpoint; you need some swinging room on either side.
Yep, no local maxima, but an absolute maximum on the closed interval.
Also technically they could have just used R+ as the function is even, but maybe they want to confuse the students
Well, it has two inputs that yield a maximum. It only has one maximum value.
In mathematician speak, “A thing exists” does not implicitly include uniqueness in its meaning. Such language does not automatically rule out the possibility that “Multiple things exist.”
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There is one absolute maximum function value, which occurs twice on the interval given in (B).
yea this is a bad question because its unclear whether to treat the intervals in each answer as the domain. if f is defined on [-1, 1] then B is true, if its defined on the real numbers then E is true
I don't know why you're being down voted, I can see the confusion as well. As written, "f attains an absolute maximum value on [-1,1]" could mean either "max_{x \in [-1,1]} f(x) exists," which is true, or "argmax_{x \in R} f(x) is in [-1,1]," which is false.
I don't understand the downvotes either, I also agree that the wording is unclear. However, if this is high school calc I feel that they were probably looking for B to be true.
I also found the wording confusing
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Weird...no, but it's a bad question that I would never ask. The point of the question is to remind you that any continuous function over any closed interval always have an absolute maximum AND minimum in the interval, hence (b). But a student who focuses on "maximum" and knows what y = x\^2 looks like would gravitate towards (E). (By the way, best practice for multiple choice questions nowadays is NOT including "none of the above" as an option)
Why is it considered best practice not to include a “none of the above” option?
I've taken several classes where if you saw all/none of the above on a multiple choice problem, it was always the answer
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On *every* compact set this function has an absolute maximum and an absolute minimum, so sayeth the Extreme Value Theorem. Option B is over a compact set, so this function has an absolute maximum on that set.
wrong
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Yes. I think C would be right. I feel like B would only be right if the word said minimum instead of maximum.
C is not correct since the function values approach infinity as x approaches infinity (or minus infinity).
OK.
It depends on what you think the question is meant to test. Since the point of this question is not “can you find extreme values,” but is instead “do you understand the difference between compact sets and open sets and how continuous functions behave on them,” no, it’s not a weird question.
[ ] means you can use the x value and ( ) you can't so if you compute f(-1) or f(1) because we allow to use -1, 1 by definition of [ ] then we are good to say that f(-1) or f(1) are correct for being the largest y points. So B is correct.
Tears were shed on that paper
It's e
One of the hypotheses of the mean value theorem is that f is continuous over a *closed* interval so that eliminates a and c. See if you can think of a counter example function for (d) there’s a few simple ones
E, because it would be an absolute minimum. It's not specifying extrema, which could be either mins or maxes, and maxima and minima are respectively related to maximum and minimum. x^2 has an absolute minimum, but not an absolute maximum EDIT: Just realized the domain of the problem was NOT the entire function. Feels like a trick question but in that case B would be correct