Can someone please explain the solution to this functions question?

[Raanikeri]

I don't understand why they let y=x^2 + 1 in the first place. And why is it later f(y-2) = (y-3)(y+1) ==>Why did they substitute f (y-2) back into the f(y) function? Isn't it usually the case to substitute the f(y) into the f(y-2)? Why would they now substitute the f(y-2) back into the f(y) to get (y-3)(y+1)? Very confused. Please help...

 

[blamocur]

I don't understand why they let y=x^2 + 1 in the first place.

The only answer which comes to mind is "because it works". Since there is [imath]f(x^2+1)[/imath] it seems natural to try replacing the argument of [imath]f[/imath] with a simple variable.

And why is it later f(y-2) = (y-3)(y+1) ==>Why did they substitute f (y-2) back into the f(y)

Because [imath]y-2 = x^2-1[/imath].

 

[Raanikeri]

The only answer which comes to mind is "because it works". Since there is [imath]f(x^2+1)[/imath] it seems natural to try replacing the argument of [imath]f[/imath] with a simple variable.

Because [imath]y-2 = x^2-1[/imath].

I guess I must not understand functions as well as I thought.

You see, all I've ever come across so far are instances where they give you the f(x) function, for example, and then ask you to find the f(x-1) function, say... So then you'd substitute f(x) into f(x-1) to get x. Like below. I've never come across any questions where you have the f(y-2) and then substitute f(y-2) into f(y), which is what was done in the solution to the original question I was asking about.

Can you confirm if this is normal - to substitute f(y-2) into f(y)? And not just f(y) into f(y-2)? The original question I asked was from a past "Maths Olympiad" type paper question so I guess it is quite outside of what we're normally tested in school. But still I'd like to improve my understanding through looking at questions and solutions to that sort of paper...

 

[Dr.Peterson]

I don't understand why they let y=x^2 + 1 in the first place.

Because they wanted to know what the function f is for any argument, so they gave a name to the argument they know, x^2 + 1, to see what would happen. (A lot of mathematical thinking involves trying something without being sure it will work -- just taking one step and hoping you've see better from the new location.)

And why is it later f(y-2) = (y-3)(y+1) ==>Why did they substitute f (y-2) back into the f(y) function?

I wouldn't describe it that way. They substituted y-2 for y in the new definition of f, because the goal was to find f(x^2 - 1), and x^2 - 1 is (x^2 + 1) - 2.

They could have done the same thing just by replacing y with x^2 - 1 literally, saying f(x^2 - 1) = (y - 1)(y + 3) where y = x^1 - 1, namely ((x^2 - 1) - 1)((x^2 - 1) + 3) = (x^2 - 2)(x^2 + 2).

I've never come across any questions where you have the f(y-2) and then substitute f(y-2)

Can you confirm if this is normal - to substitute f(y-2) into f(y)? And not just f(y) into f(y-2)?

Again, you are describing this in a way that may be confusing yourself. The expression f(y-2) just says, what comes out of the "machine" when you put y-2 into it.

The original question I asked was from a past "Maths Olympiad" type paper question so I guess it is quite outside of what we're normally tested in school.

Olympiad type questions are probably all about turning things inside out to challenge you with non-routine thinking!

 

[Raanikeri]

Because they wanted to know what the function f is for any argument, so they gave a name to the argument they know, x^2 + 1, to see what would happen. (A lot of mathematical thinking involves trying something without being sure it will work -- just taking one step and hoping you've see better from the new location.)


I wouldn't describe it that way. They substituted y-2 for y in the new definition of f, because the goal was to find f(x^2 - 1), and x^2 - 1 is (x^2 + 1) - 2.

They could have done the same thing just by replacing y with x^2 - 1 literally, saying f(x^2 - 1) = (y - 1)(y + 3) where y = x^1 - 1, namely ((x^2 - 1) - 1)((x^2 - 1) + 3) = (x^2 - 2)(x^2 + 2).



Again, you are describing this in a way that may be confusing yourself. The expression f(y-2) just says, what comes out of the "machine" when you put y-2 into it.


Olympiad type questions are probably all about turning things inside out to challenge you with non-routine thinking!

Alright I think I'm starting to understand it a bit better now, when you say think of it as machines... so one machine processes y-2, the other machine processes y, but you can put either machine into the other? I hope I'm on the right track...
Is this question a form of iteration?

 

[lev888]

Alright I think I'm starting to understand it a bit better now, when you say think of it as machines... so one machine processes y-2, the other machine processes y, but you can put either machine into the other? I hope I'm on the right track...
Is this question a form of iteration?

There is only one machine - the function f. We are not "putting one machine into the other". We are putting different inputs into the same machine.

 

[Raanikeri]

There is only one machine - the function f. We are not "putting one machine into the other". We are putting different inputs into the same machine.

Okay I see..
I notice function f(x) was never found or defined in this question. Would it be possible to find what f(x) is equal to?

 

[Dr.Peterson]

Okay I see..
I notice function f(x) was never found or defined in this question. Would it be possible to find what f(x) is equal to?

No, f(x) was found! Keep in mind that the variable used is just a place-holder; if you know that f(y) = (y-3)(y+1), then you also know that f(x) = (x-3)(x+1).

This is an essential part of understanding function notation.

 

[Raanikeri]

Ooh! I see! I don't think I'd ever realise this if I hadn't tried to do this question and asked on here. So they're trying to

No, f(x) was found! Keep in mind that the variable used is just a place-holder; if you know that f(y) = (y-3)(y+1), then you also know that f(x) = (x-3)(x+1).

This is an essential part of understanding function notation.

I understand the solution now, thank you for your patience...

But this last bit you said has baffled me a little. You say f(y)=(y-3)(y+1), therefore f(x)=(x-3)(x+1). But isn't y=x^2+1? so how can y = x now? x^2+1 is not equal to x, right?

 

[lev888]

But this last bit you said has baffled me a little. You say f(y)=(y-3)(y+1), therefore f(x)=(x-3)(x+1). But isn't y=x^2+1? so how can y = x now? x^2+1 is not equal to x, right?

f(y)=(y-3)(y+1) means that for any valid input y the result of the function is (y-3)(y+1). We can name that input y, x, ABC, cow, etc.

 

[Dr.Peterson]

Ooh! I see! I don't think I'd ever realise this if I hadn't tried to do this question and asked on here. So they're trying to

I understand the solution now, thank you for your patience...

But this last bit you said has baffled me a little. You say f(y)=(y-3)(y+1), therefore f(x)=(x-3)(x+1). But isn't y=x^2+1? so how can y = x now? x^2+1 is not equal to x, right?

As I said, the input variable in a function definition is a placeholder, also called a dummy variable, parameter, or argument. It just means "whatever is taken as the input".

Given, say, the definition f(x) = 2x - 3, we can apply it to any input we want by replacing x with a specific value:

• f(4) = 2(4) - 3 = 5
• f(x+1) = 2(x+1) - 3 = 2x - 1
• f(y) = 2y - 3

and, yes,

• f(cow) = 2(cow) - 3

(Well, actually, "cow" is not in the domain of this function ...)

See

What is a Function

 

[blamocur]

(Well, actually, "cow" is not in the domain of this function ...)

How is "cow" different from "y" ? Only because mathematicians prefer single-letter variable names (usually meaning different things depending on the type face, and often adorned with bunches of sub- and sup-scripts), but in software domain no respectable compiler would notice any difference