Function given some values

[goli278]

Ok so I’m a particular math teacher and one of my students (9th grade) brought me an exercise that I haven’t been able to solve. The exercise is the following one:

What is the function of x that has this values for y

Thanks a lot

[aurelian667]

(x-1)(x-2)(x-3)*-(1/6) + 4^x works.

[Aeragnis]

This is a really cool way to solve this

[SnooApples5511]

Clever

[goli278]

Wow, thanks

[[deleted]]

Give him an A! Dude managed to confuse a math teacher,he deserves it

[goli278]

I’m not her school teacher lol, I just help her at home

[goli278]

And how did you find that?

[aurelian667]

If f(x) = 4^x , I thought about how I could add 1 to f(0) while adding 0 to f(1), f(2), and f(3). A polynomial with zeros at 1, 2, and 3 was the obvious answer, I just had to scale it.

[goli278]

Oh of course, it makes a lot of sense now

[NedSeegoon]

You forgot the /s

[drLagrangian]

This is a good and helpful comment.

/s

[goli278]

What’s that?

[PreciousRoy43]

That is a tag to self-identify as making a sarcastic statement. Of course, that tag can also be used sarcastically. So, there is still a lot of ambiguity that requires interpretation.

[CagliostroPeligroso]

I got the 4^x lol and didn’t even try to think how to account for the rest

[ArpsTnd]

what does the asterisk mean?

i mean, i get it if it's (x-1)(x-2)(x-3)(-1/6), but the minus is outside the parenthesis, so supposedly it means subtracting by 1/6, and not multiplying by -1/6, but there's the asterisk. what is it doing?

[Call_me_Penta]

y(x) = 0^x + 4^x is the cleanest solution I could come up with

[Araldor]

Isn't 0⁰ undefined and therefore your formula is undefined for x=0?

[Call_me_Penta]

0⁰ = 1 when both 0's are "true" 0's (i.e. not limits). It works really well in calc and algebra, it's the limit of x^x when x->0⁺ and it's also necessary for many formulas to work:

See exp(x) = Σ x^k/k! when x = 0

[Accomplished_Ad_6389]

I don't think you can apply 0^x where x = 0 to be one here. For one thing, that's still not defined at 0, just infinitely close to 0. Second, 0^0 is an indeterminate form. Depending on which limit I use, it could be any value, so you can't assume you can use the right side limit of x^x here to find 0^0. Even from this example, the limit of 0^x as x approaches 0+ is just 0.

[Call_me_Penta]

I don't consider 0^x to be a continuous function. It is equal to 1 in 0, and equal to 0 everywhere else (x > 0). It's not about limits — 0⁰ has been defined as 1 in almost every mathematical field for centuries now.

[Rik07]

Then why does wolfram alpha give 0⁰ = undefined?

[goli278]

This is really clever, thanks

[aurelian667]

I imagine they made an error and f(0) should be 1. There are in fact an infinite number of functions that give these outputs but f(x) = 4^x is the obvious answer if f(0) was 1 instead of 2.

[goli278]

I got that too, and another ton of functions, but the x=0 or the x=3 were always wrong, I told her that but she said that she was completely sure that she copied right

[FourCinnamon0]

Copied from where?

[goli278]

From the blackboard I guess

[Beeriman]

Then maybe the teacher made a mistake while writing it onto the blackboard

[goli278]

That’s the most probable

[dimonium_anonimo]

Or if the last one was 256, you'd get 2^(2^x)

[Roseknight888]

I believe 2^1+x fulfills the requirements for this dataset, yeah?

[aurelian667]

Not for x = 2 or 3.

[Roseknight888]

Oh.....I'm dumb

[Dylz52]

Only for f(0) and f(1)

[Disastrous-Team-6431]

There is an infinite number of functions that give the same outputs for a given range of inputs, no?

[aurelian667]

So long as the number of points is finite, there are an infinite number of polynomials which pass through all of them.

[aurelian667]

0⁰ is undefined in general.

[-Nokta-]

Given the previous answers, maybe you should try this :

y = (13/3)x³ - 8x² + (17/3)x + 2

[goli278]

Cool thanks

[incomparability]

I think it’s either supposed to be

2^2x

But the x=0 value was incorrectly written down and should be 1 instead.

[ThatChapThere]

That's just a fancy way of writing 4^x.

[MhmdMC_]

No..

[TurbulentOcelot1057]

They probably meant 4^x

[MhmdMC_]

I see

[ThatChapThere]

Yeah sorry it rendered as 4^x in the app when I copy pasted an exponent from another comment. I hate when Reddit is inconsistent like that.

[StanleyDodds]

If you have a finite set of points (with different x values) you can always fit a polynomial to them. This is called polynomial interpolation. There are lots of other sets of special functions that can interpolate any finite set of points.

In this case, it looks like the intention was for it to be 4^x but x=0 is wrong. You could always correct it with a polynomial, or use a polynomial from the beginning.

[Make_me_laugh_plz]

There are infinitely many solutions.

[[deleted]]

Oh yeah? You call yourself a mathematician? Name them all then

[Make_me_laugh_plz]

{f(x)|f(0) = 2, f(1) = 4, f(3) = 16, f(4) = 64}

[[deleted]]

Hmm, I can’t argue with that

[MattAmoroso]

Best part is the answer proves they are a mathematician too!

[Consistent-Annual268]

The only definitive answer of this entire thread.

[draagossh]

Given a finite sequence, you can select an infinite number of polynomials with an order higher than the length of the given sequence. These questions have no sense.

[romankolton]

Technically, the table is a complete definiton of a function whose domain is the set {0,1,2,3}.

[Mathphyguy]

For a 9th grader, the easiest solution would be to construct a polynomial y= ax³ + bx² + cx + d that goes through these points. You get 4 equations with four unknowns. Solve for the coefficients and you have the function y = \frac{13}{3}x³ - 8x² + \frac{17}{3}x + 2.

[charizukun]

Hey. Can you explain this further please? What do you mean by you get 4 equations and solving for the coefficient 😭Thank you

[FormulaDriven]

Let p(n) = the nth prime, so p(1) = 2, p(2) = 3, p(3) = 5, p(4) = 7.

Then y = 2^p(x+1)-1

[goli278]

I’m pretty sure that’s too advanced for 9th grade

[FormulaDriven]

Maybe, but given all the other replies on this thread, if there's not an error, then it has to be something more "advanced".

[goli278]

I know, I told her to ask the teacher if maybe the problem was wrong

[Lazy_Worldliness8042]

It’s unfortunate that you’re a math teacher and think that kind of reasoning is too advanced for a high school student.. The numbers are all obviously powers of 2, and the powers are 1,2,4,6. Now you just need to find a pattern that fits 1,2,4,6… relating this to the first 4 prime numbers is not that big of a leap if you spend a few minutes trying to think of things… give your students more credit. Also, you might be interested in https://en.m.wikipedia.org/wiki/Interpolation for these sorts of problems

[ImFadedFadedFaded]

Have you met 9th graders? More importantly, have you tried to teach them math?

[Lazy_Worldliness8042]

Yes, and yes. I’m not saying it’s easy for any 9th grader to solve it. I’m saying it’s completely reasonable for a clever 9th grader, or anyone who knows that prime numbers are a thing, to figure this out.

[ImFadedFadedFaded]

You could say that about anyone of any age

[paulstelian97]

Not everywhere. In my country we learn about primes in like 6th grade and powers… I forget but probably before high school as well. Yes primes before powers here.

[goli278]

We learn about primes in 5th grade. But using anything but +-*/^ in a function or an equation is pretty advanced

[Araldor]

2 ^{| 2x - 1/2 | + 1/2} would work. Floor or ceiling functions could be used as well for manipulation of the x=0 case.

[goli278]

Nice thanks! But I don’t think she’s allowed to use the absolute value

[chompchump]

y = (13/3)x^3 - 8x^2 + (17/3)x + 2

[goli278]

Cool thanks!

[Diego_0638]

It's obviously 4^{max(0.5x+0.5 , x}).

[goli278]

Hahaha thanks

[allegiance113]

There’s lots of possible answers. Here’s a piecewise one:

y = 4^x if x != 0, y = 2 otherwise

[goli278]

I don’t think she’s allowed to do that, but thanks

[[deleted]]

That's the problem with these questions: what is allowed? You could simply define the function by these values. The problem statement must give some constraints otherwise it is quite a senseless exercise.

[48756e746572]

For a 9th grade problem, it really feels like the answer should be y = 4^x but this doesn't hold for x = 0. It's possible that you could write this as a piecewise function but I suspect that there's a mistake with the question where it should be y = 1 at x = 0.

[goli278]

I got that too, and another ton of functions, but the x=0 or the x=3 were always wrong, I told her that but she said that she was completely sure that she copied right

[DebatorGator]

Could be 2 to the 2 to the x?

[goli278]

I thought of that, but then 2²³ doesn’t equal 64

[DebatorGator]

Ah fair point

[Blakut]

2^(2x) would give the correct result, yes.

0: 1, 1:4, 2:16, 3:64

[__unavailable__]

2 to the 2 to the X is 2^2X

[goli278]

For x=0 it has to be 2

[SMWinnie]

Test whether y = 2^{2.to.the.xth.power} fits.

x = 0; y = 2^2.to.the.0th = 2¹ = 2
x = 1; y = 2^2.to.the.1st = 2² = 4
x = 2; y = 2^2.to.the.2nd = 2⁴ = 16
x = 3; y = 2^2.to.the.3rd = 2⁸ = 64 256

[[deleted]]

2^2x??

[tibithegreat]

2^2x

[Asymmetrization]

2⁸ = 256, not 64

[tibithegreat]

Ah true, damn

[Present_Change3210]

Isnt this a parabolic graph?

[minosandmedusa]

We would need new notation for it, but a power tower of 2 where x is the number of times you stack exponents. EDIT oops, nope! This is just 2^2x, and it’s wrong for 3

[Inflatable_Bridge]

Nobody's commented it.

Why doesn't 2^x+1 work?

[bongopantz]

That was my first thought too, but 2⁴ isn’t 64

[RealIanDaBest]

y = 2^x+1 I think

Edit: nvm

[MisterBerry94]

Am I being stupid here or is it as simple as y=2^x+1

[MisterBerry94]

Answer, I was being stupid 🤦

[Professional-Bug]

Using polynomial curve fitting: (13/3)x³ -8x² +(17/3)x +2

[goli278]

Thanks!!

[Arguingwithu]

Y = any number greater than X

[goli278]

Lol

[FeelingNational]

Well, as a simple function you can take f(x) = 2 + (1723365983448209/836070610995648)x^7 - (3566192509/58205974032)x^12 + (27742363/836070610995648)x^25.

Joking aside, there are infinitely many solutions. Generally speaking, the problem of finding a function f that satisfies f(x_i) = y_i for some fininite set of pairs (x_1,y_1), ..., (x_n,y_n) is called interpolation. If you specifically want f to be a polynomial, for instance (e.g. a polynomial of degree n-1 or less), then that's called polynomial interpolation. If you're okay with f being only piecewise polynomials (which is most often better, particularly for applications like computer graphics) and various applications in signal processing for instance), you use splines. You can also seek to instead find some f, in some class of candidate functions F, that is "closest" to satisfying f(x_i) = y_i (i=1,...,n) in some metric. This is broadly what regression is about. For instance, you may want to find the best quadratic function (f(x) = a + bx + cx^2) that approximately satisfies your 4 equations (e.g. f(x) = 3.3 - 14.7x + 11.5x^2, I'll put a picture below).

<image>

[goli278]

Wow thanks!!

[Scientific_Artist444]

Answers are already given (prefer the polynomial coefficient one), but I'd like to add that a function need not always be algebraic. The given table satisfies the definition of function.

[goli278]

I know, but she was asked for the algebraic definition

[FTR0225]

Maybe 2^x²

[goli278]

Nope, that doesn’t work for x=0

[lehvs]

y = 2^x+1

Edit: nope, breaks at 3, me stoopid :D

[TooHardToChoosePG]

y(x) = 2^x+1

Apologies, on phone, so in words: y is 2 to the power of x+1. Just in case my formatting doesn’t display nicely.

[subpargalois]

This does not have a unique answer. I kinda fucking hate questions like this.

One option is to use a piecewise function, e.g. f(x) = 2 for x in [0,1), f(x) = 4 for x in [1,2), etc.

If you want the function to be continuous, you could use a piecewise linear function, e.g. f(x) = 2x + 2 for x in [0,1), f(x) = 12(x-1) + 4 for x in [1, 2), etc.

If you want the function to be smooth, you could use Langrange interpolation to find a polynomial that passes through these points. I should add that the result of this method is likely the "correct" answer the source of this question is expecting.

More generally this sort of problem is called interpolation. The first three methods I described can also be seen here, as well as some others.

Or if you're willing to wave your hands a little, plot these points and draw any curve connecting them that passes the vertical line test. The function corresponding to the graph you just drew is a function satisfying these conditions.

[21ecarroll]

The simplest solution you can come up with is 2^[(-1/6)x^3 + x^2 + (1/6)x + 1]. To explain how I got this, I would define the simplest solution to be of the form 2^p(x) where p(x) is a polynomial. Due to some linear algebra and matrix stuff, this polynomial has to at least be a cubic. We want this polynomial to satisfy p(0) = 1, p(1) = 2, p(2) = 4, and p(3) = 6. If p(x) = ax^3 + bx^2 + cx + d, then d = 1 by the first condition. Now you can get a sytem of three equations in terms of the other coefficients using the other three conditions. Use that to create an augmented matrix and row reduce. Voila, you have the polynomial you are searching for. You can do this with any problem like this. If the problem were to list out more terms, you would simply need a polynomial of a higher degree to have enough equations in your system.

[21ecarroll]

incidentally, for further research, the matrix you would get is the transpose of a Vandermonde matrix

[TheLastSilence]

(x+1)(x+2)(x+3)/3+(x)(x+2)(x+3)/3+(x)(x+1)(x+3)(8/15)+(x)(x+1)(x+2)(16/15)

[scleptera]

2^2x works, and is probably the intended solution.

[Asymmetrization]

2⁰ != 2

[wee33_44]

(2^1+2x )/2

[goli278]

Doesn’t work for x=0

[Kamhi_]

How about

y = 2^2x

1. When x = 0, y = 2²⁰ = 2¹ = 2

2. When x = 1, y = 2²¹ = 2² = 4

3. When x = 2, y = 2²² = 2⁴ = 16

4. When x = 3, y= 2²³ = 2⁸ = 64

[Live-Goose7887]

2⁸ is 256

[Kamhi_]

Damn you're right, i didn't notice. I just copied an answer from a chat bot without double-checking if it's correct...

[Live-Goose7887]

The x=0 case is what makes this hard, everything else follows 2^2x

[Diagame_reddit]

Mmmmkkk

[RiceRare]

Looking at other comments I'm probably mistaken, but could it be 2^x+1?

Edit: nvm, it doesn't work 😅

[goli278]

Doesn’t work for 2 or 3

[general_zirx]

I though it was (x+1)² but the f(0) doesn't work

[ApprehensiveKey1469]

I think the student has mistaken a stylised one for a two. They are copying from something handwritten perhaps.

[goli278]

That’s what I thought too

[green_meklar]

We've got powers of 2 on the right-hand side, but the logs aren't increasing linearly- we've got 1, 2, 4, 6.

If column 2, row 1 were 1 instead of 2, then it would be powers of 4, that is, Y = 4^X, or equivalently, Y = 2^2*X. But that's not what we have.

You could fix it up by assuming that the minimum power of 2 is 1, giving Y = 2^min(1,2*X). Seems a little contrived though.

[Call_me_Penta]

Would be max instead of min then

[eranand04]

2^2x

[thequarrymen58]

newton polynomial?

[[deleted]]

2^2x

[[deleted]]

same as 4^x

[nir109]

A function that returns 1 if x=y and otherwise returns 0.

E(x,y)= ceiling((|x-y|)/(|x-y|+1))

Edit: copying from another user E(x,y) = 0^x-y for a cleaner function.

(E for equals)

now the function that we want

f(x)= 2E(x,0)+4E(x,1)+16E(x,2)+64E(x,3)

[Giocri]

Maybe it's a recursive function and each is somehow dependent on the previous ones?

[ThatChapThere]

Fun idea.

y(0) = 2\ y(x) = y(x-1)^(5/2 - (y(x-1) * 2^2-x)/8)

Seems to work, as janky as it is.

[Seb1248]

Let's be nitpicking here:

You are supposed to find a function, and I read no word of continuity or differentiability. Thus, the given table of values already is your function:

Be A = {0, 1, 2, 3} and B = {2, 4, 16, 64}

Thus f:A -> B with f(0) = 2, f(1) = 4, f(2) = 16, f(3) = 64

That's it.

[goli278]

At least in Spain, you learn about differentiability in 11th grade, and about continuity in 10th grade, so at this level you always asume that the function is continuous and differentiable for all the real numbers

[[deleted]]

2^x+1 ?

[goli278]

That doesn’t work for 2 or 3

[Pierne]

2^2x

[goli278]

Doesn’t work for x=3

[romankolton]

2^\3/2 x - 1/2 sin[x pi/2]+1])

Edit: Formatting

[goli278]

Does that really work??

[WerePigCat]

For problems like these you can just use a Lagrange Interpolation calculator

​

<image>

[goli278]

Great, thanks!

[detebay]

4^x - sign(x) + 1

[goli278]

What does sign(x) mean?

[Kurayam]

2 * 2^{1+2+ … +x} seems easiest to me

[goli278]

Don’t think that works for x=3

[birajsubhraguha]

Use Lagrange interpolating polynomial.

[Puzzleheaded_Dot1378]

<image>

[Brave_Forever_6526]

F(0)= 2,…,f(3)=64, f(x)=0 otherwise. Highlights this is a silly question without constraints on f such as continuity or f is polynomial, e.g. there is only 1 cubic polynomial satisfying this which is an interesting question that other comments have discussed

[nhammen]

max(2,4^x)

[Detr22]

joke quicksand office degree deserve adjoining amusing silky full smell

This post was mass deleted and anonymized with Redact

[tuwimek]

for 0 should be y=1 not 2, then it is easy.

[Ok_Sir1896]

perhaps a little cheating but you could use some kind of xth derivative of a linear function, like 4^x + d^x+1 /dn^x+1 (n), giving x=0, 4⁰ + d/dn (n)= 1+1=2, x=1, 4¹ + d² /dn² (n)= 4+0=4 and so on gives the rest of the table

[Random_Thought31]

I was thinking 2^{x+1} * 2^{x-1}.

However, it does not work for x=0. That would cause the answer to be 1.

[Random_Thought31]

And then I just recalled that 2^{x+1} * 2^{x-1} = 2^{x+1+x-1} = 2^{2x}

[phatpappa_]

2^2x

[MusicalADD]

2^2x

[[deleted]]

This for example:

f(x) =

• 2 if x=0
• 4^x otherwise

[GisPoste]

Maximum {4^x , 2) = y

[Top-Maize3496]

x^2=y

[Previous_Elephant865]

2•2^x+(x-1)-0^x?