Please explain how functions work. I watched all the videos on PrepSwift. I tried to go through the problems. I read through your explanations. And I want to throw my computer across the room, because no matter how hard I try, I can’t understand how 5 can equal 8.
Let’s take your example:
f(x) = 1/f(x-1) and f(3) = 1/2
What is greater, f(23), or 1/2?
To me, this registers as: 3 = 1/??? something with 1/2. So 3 = 1/(1/2 - 1) → 3 = 1/-0.5 → 3 = -2? Or am I supposed to imagine for every 3 put into something, it has this output that also has a 1/(1-2 - 1)??? I’m so confused by “every time you do ___, you get ____.”
Another example: f(x) = f(x+3) and f(5) = 10. You video says, 5+3 = 8, then says that it’s the exact same when it’s f(8) = 10, f(11) = 10, etc. This is so confusing, because it looks like for every 5, you’re supposed to have 10. But how can 8 = 10? If I plug in f(8) —> f(8+3) = 11, how does 8 = 11???
I don’t understand the terminology of inputting something and using the same f(x) for both the input and output sides with these stupid f and (x) signs. I have no idea what’s going on. Literally every other video up to here has made sense in the arithmetic and all other algebra. What am I missing here??? I’m at 90+ minutes in on this and it won’t sink in.
It can help to take a “first-principles” approach to this problem. Let y = f(x). Based on your understanding of functions till date, what do you think that equation means? Explain as well as you can. Don’t worry about the correctness or quality of your answer at this stage.
If y = 2, that means that y is equivalent to integer 2.
So in my head, if y = f(x), that means y is equivalent to whatever f(x) is. F(x) in my mind is some sort of variable.
But apparently that’s wrong because it’s an input … or an output, depending on where it is relative to the = sign. I don’t understand what that means, apparently. Or if you say:
F(5) = 10, this to me is confusing. How can 5 = 10? Or assuming that when “input” is 5, Y = 10.
So for your y = f(x), if you say F(5) to me, that means y = 5. But apparently that means it actually equals 10.
Consider a black box. You feed in an input, it does something (what that is you may or may not know), and spits out an output.
So if we go back to the original example - we feed in x, the black box does something, and returns y. The “black box” is essentially the function here.
If f(5) = 10, this means that the black box (i.e, the function) is given an input 5, and does something to it (it could be doubling the input, adding 5 to it - we don’t know unless told so explicitly), and spits out the output 10. Does that help?
Yes, that makes sense, but I don’t understand how it works.
Can you illustrate that with the original two problems I posted? And explain along the way.
I also don’t understand what the input VS machine/function VS output is.
If the input is the 5, the output is the 10, is the “f” the machine/function? The symbols do not make sense to me. As in f(x) is a “function”, but you can put in an “x,” or a “5.” So if f(x) is a function and you input an “x,” is that both a function and an input? Why do I need to have an “f” at all?
I need this explained like I’m five.
We’ll get there.
Yes.
So the black box needs an input. This can indeed be x or 5 (or whatever).
x → the input
f(x) → the function (aka the blackbox).
The output of this function is something else - which we usually indicate with another variable. So in y = f(x),
- x → input
- f(x) → the black box doing something to the input and returning something else
- y = the output.
If y = f(5),
- 5 → the input to the black box
- f(5) → the black box doing its magic
- y → the output
It may also help to visualise the problem in a flowchart-like manner:
Can you hence understand the intention behind this diagram?
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I get the black box illustration (before you put the numbers in). But it seems like the terms overlap.
For example, is f(x) its own term? Not the same as “x”?
In your example, you put 5 in, and you get 5 out. f(x) changes into f(5), which is fine - that looks like you’ve chosen 5 as your “x” value or input.
I don’t understand how the term changes from f(x) to f(5) if f(5) is also the output in your picture.
Basically, everything looks like it’s 5. Here’s what I see from your picture:
5 (x or input) —-> f(black box) which is also f(5) or f(x)? —-> f(5) (y or output)
Perhaps the second example in my OP world be more helpful since everything is not equal to 5?
f(x) is just maths for “feeding x in the black box, doing something and returning some output”. So f(x) is putting x as the input, while f(5) is putting 5 as the input (and getting some output).
Does that help?
Yes, thank you. Can you go through the original 2 examples I posted?
If we go to the first problem:
This is similar to the initial examples I provided, but you’re given some information about the “black box”. In this case, that f(x) = \frac{1}{f(x - 1)} and f(3) = \frac{1}{2}. The goal of this question is to determine, using the hints that you’ve already been given, what the black box would output if it’s given 23 as the input. Can you do it?
No, I want you to show me. I already did the problem on prep swift, and don’t understand where the inputs are going. I know it’s looking for even and odd patterns.
If I input in 1/2 on the right side of the equation for f(3), it makes sense that 1/2 - 1 = -(1/2). But I’m not doing anything to the left side other than changing the input.
So for function of 3, input of 1/2 makes the right hand side of the equation to be 1/-(1/2) = -2.
So in this case, I just plug in the 1/2 into the “x” value on the right hand side of the equation.
But I don’t understand where suddenly the f(4) comes from and how putting in a 4 can still be the same as putting in a 3 and getting 1/2.
I don’t understand where the information is coming from that tells me when I input “4” instead of “3” what kind of an effect it’s going to have. I can’t just assume that f(4) = 2 or w/e it is, because that information hasn’t been given.
Where are you getting the -1 from?
Actually the information has been given - just in a scaffolded form here. That’s the challenge.
The -1 is in the denominator where it says “f(x-1).”
If f (3) = 1/2, doesn’t that mean that 1/2 is plugged into the “x” value?
Which would be f(1/2 - 1)?
Sorry, this is why I am confused; I don’t understand where to plug things in.
No. Recall from the beginning what the notation means. If y = f(x), x is what we’re feeding into the function and y is the resulting output. So, in f(3) = \frac{1}{2}, 3 is the input, the function (i.e, black box) does something), and \frac{1}{2} is the output. So 3 is what is plugged in, not \frac{1}{2}.
So in the above question, the equation
f(x) = \frac{1}{f(x - 1)}
is essentially providing you with a hint on the black box. That is, the output when x is input is the same as the output when x - 1 is input. So if x = 4, this becomes
f(4) = \frac{1}{f(3)}
Because we already know what the output of the function when 3 is input, and the “hint” about the black box that we were given earlier, we can find out the output when 4 is input. Can you do that?
Does this help? I know that this is slow, but it is important that we get the foundation right as that will help significantly in solving the other problems.
Sorry, I fell off here for awhile.
Can you just step by step solve the problems and a few other examples? I don’t want to solve them, because I still don’t visualize it well.
If I can see you solve several, that will help me to plug things in correctly and connect the dots.
We don’t really require a PERFECT foundation. We usually want users to have what we call an 80% or 85% foundation. Whenever we try to be perfect in our foundation, we never focus on strategies, untimed practice, and timed practice. So I guess my question is does it make more sense for you to skip functions?
Perhaps you’re right. I think I’ll look at the examples in the 5 lb Book and see if I can figure it out, or jettison it otherwise. Thank you.
