I need help understanding fuctions.

Jack Omar · 2023-01-16 07:43:28

Hello !

I am having problem understanding the functions or symbols I am not sure.
For example let's say

this is my given function, then this symbol is again used and then it is interpreted as . I don't get it, it also can be a function (a constant function), so why do I replace with number 2 ? Why doesn't it overwrite the function. What do I mean by overwrite ? When is a linear function so then I change it into which is constant. Why it is not overwriting the function when I used the same symbol which is ?

The similar problem appeared in derivative for example :

and the same symbol but with different formula so now I can have . I don't get it. So when I have this : , f'(x) is a derivative of a function, but then I have the same symbol but with different formula : , doesn't this change how I calculate the derivative f'(x) I just now calculate the derivative using the second formula.

I see similarities in both situations but I don't get it, when I am calculating the f(x) or when I have a result of f(x) or f'(x). Can someone guide me please

Bob · 2023-01-16 22:48:19

hi Jack

I made this picture some years ago for someone learning about functions.

A number goes into the box; the function makes a new number come out. (You can also have more than one input but let's not go there at the moment!)

Mathematicians often abbreviate the function box to f(x), meaning f is the box, x in the input.
And if you know f(7) = 5 for example, then 7 is the input and 5 is the output.

So in your example, f(x) = 4x -2 you are being told that the rule for what to do with the input is "multiply the input by 4 and subtract 2; output the answer".

So f(7) = 4 x 7 - 2 = 26

f(-2) = 4 x -2 - 2 = -10

If the questioner wants a second function they'll use another letter so the two don't get muddled.

eg. g(x) = 8x so g(5) = 40

You say you have two ways of showing differentiation. Actually only one of these is correct:

This comes about by a procedure known as differentiating from first principles. This is the procedure:

I'll put my coordinates (x,y) in square brackets [x,y] so you don't get muddled with the function brackets.

1. Pick a point on the curve [ x, f(x) ]

2. Pick a second point close to the first [ x+h, (f(x+h) ]

3. Work out the gradient of the line joining the points:

4. Let h tend to zero and simplify the expression. Because the line joining the points gets closer and closer to the gradient at the first point, we conclude this is the gradient at the point. One way of writing the gradiant function is to put a ' after the f = f'

The second expression:

is just an expression for the gradient of a line joining [ a, f(a) ] to [ b, f(b) ]

It won't be the gradient at a point until you have applied the first principles rule.

Would you like some examples of how the 'first principles rule' works in practice?

Bob

Jack Omar · 2023-01-21 05:29:34

Bob wrote:

hi Jack
I made this picture some years ago for someone learning about functions.
https://i.imgur.com/gQXbI23.gif
A number goes into the box; the function makes a new number come out. (You can also have more than one input but let's not go there at the moment!)
Mathematicians often abbreviate the function box to f(x), meaning f is the box, x in the input.
And if you know f(7) = 5 for example, then 7 is the input and 5 is the output.
So in your example, f(x) = 4x -2 you are being told that the rule for what to do with the input is "multiply the input by 4 and subtract 2; output the answer".
So f(7) = 4 x 7 - 2 = 26
f(-2) = 4 x -2 - 2 = -10
If the questioner wants a second function they'll use another letter so the two don't get muddled.
eg. g(x) = 8x so g(5) = 40
You say you have two ways of showing differentiation. Actually only one of these is correct:
This comes about by a procedure known as differentiating from first principles. This is the procedure:
I'll put my coordinates (x,y) in square brackets [x,y] so you don't get muddled with the function brackets.
1. Pick a point on the curve [ x, f(x) ]
2. Pick a second point close to the first [ x+h, (f(x+h) ]
3. Work out the gradient of the line joining the points:

4. Let h tend to zero and simplify the expression. Because the line joining the points gets closer and closer to the gradient at the first point, we conclude this is the gradient at the point. One way of writing the gradiant function is to put a ' after the f = f'
The second expression:
is just an expression for the gradient of a line joining [ a, f(a) ] to [ b, f(b) ]
It won't be the gradient at a point until you have applied the first principles rule.
Would you like some examples of how the 'first principles rule' works in practice?
Bob

Hello Bob !

Sorry for not answering for so long.
I think that I understand the difference between

and . The first one says how to calculate the output f'(x) and the second one is actually the output of f'(x) but I don't know the value "x" for this f'(x).

Something like that ?

PS.
Could you also tell me how this lim works in f'(x) ? I've seen this picture :

I don't get it because

is going to 0 and is also going to zero, but somehow I got a normal value for f'(x).

Bob · 2023-01-21 07:24:08

hi Jack,

Loads of books will tell you you cannot find the value of 0 divided by 0.

Let's explore that first.

If I see the calculation 48 / 6 ; one way to answer this is to ask "What must I multiply by 6 to get the answer 48?"

This works because x and ÷ are opposite operations. This works for all divisions except one.

You cannot evaluate 0 / 0 by asking "What must I multiply 0 by to get 0?" Because any number times 0 will make 0.

This is the fundamental problem with trying to get the gradient at a point. If you pick a point A and write down its coordinates (x,y) then write down the coordinates of B where B is (x+0, y+o) you will hit the above difficulty when you try to do:

gradient = (difference in y coords) / ( difference in x coords) = ( y+0 - y ) / ( x + 0 - x) = 0 / 0

The differentiation from first principles method is a way of getting around this problem.

Let's look at this with an example.

Function: y = x^2 A = ( 3 , 9 )

Let's try B = ( 3.5 , 12.25 )

The gradient of AB is ( 12.25 - 9 ) / ( 3.5 - 3) = 3.25 / 0.5 = 6.5

Now try B = ( 3.1 , 9.61 ) G = ( 9.61 - 9 ) / ( 3.1 - 3 ) = 0.61 / 0.1 = 6.1

Now try B = ( 3.001, 9.006001) G = ( 9.006001 - 9 ) / ( 3.001 - 3) = 0.006001 / 0.001 = 6.001

You can try some more values for B for yourself. eg, B = ( 3.000001, ...........

What you will notice is that as B gets closer to A the value of the gradient, G, gets closer to 6.

You can also try B coordinates left of A such as ( 2.9, 8.41 ) You'll find the G values are under 6 but get bigger towards 6 as B moves closer to A.

Conclusion: The value of G at ( 3, 9 ) is 6.

Here's some algebra to justify this. Let B = ( x + h , x^2 + 2xh + h^2)

G = ( x^2 + 2xh + h^2 - x^2) / ( x + h - x ) = ( 2xh + h^2 ) / h

At this stage, letting h = 0 will get us nowhere as we'll end up with 0 / 0. The trick is to simplfy first.

G = 2xh / h + h^2 / h = 2x + h.

Now we can safely let h become zero as we are left with 2x. So we can conclude f'(x) = 2x

Please let me know if this makes sense so far.

Bob

Jack Omar · 2023-01-21 09:19:45

Bob wrote:

hi Jack,
Loads of books will tell you you cannot find the value of 0 divided by 0.
Let's explore that first.
If I see the calculation 48 / 6 ; one way to answer this is to ask "What must I multiply by 6 to get the answer 48?"
This works because x and ÷ are opposite operations. This works for all divisions except one.
You cannot evaluate 0 / 0 by asking "What must I multiply 0 by to get 0?" Because any number times 0 will make 0.
This is the fundamental problem with trying to get the gradient at a point. If you pick a point A and write down its coordinates (x,y) then write down the coordinates of B where B is (x+0, y+o) you will hit the above difficulty when you try to do:
gradient = (difference in y coords) / ( difference in x coords) = ( y+0 - y ) / ( x + 0 - x) = 0 / 0
The differentiation from first principles method is a way of getting around this problem.
Let's look at this with an example.
Function: y = x^2 A = ( 3 , 9 )
Let's try B = ( 3.5 , 12.25 )
The gradient of AB is ( 12.25 - 9 ) / ( 3.5 - 3) = 3.25 / 0.5 = 6.5
Now try B = ( 3.1 , 9.61 ) G = ( 9.61 - 9 ) / ( 3.1 - 3 ) = 0.61 / 0.1 = 6.1
Now try B = ( 3.001, 9.006001) G = ( 9.006001 - 9 ) / ( 3.001 - 3) = 0.006001 / 0.001 = 6.001
You can try some more values for B for yourself. eg, B = ( 3.000001, ...........
What you will notice is that as B gets closer to A the value of the gradient, G, gets closer to 6.
You can also try B coordinates left of A such as ( 2.9, 8.41 ) You'll find the G values are under 6 but get bigger towards 6 as B moves closer to A.
Conclusion: The value of G at ( 3, 9 ) is 6.
Here's some algebra to justify this. Let B = ( x + h , x^2 + 2xh + h^2)
G = ( x^2 + 2xh + h^2 - x^2) / ( x + h - x ) = ( 2xh + h^2 ) / h
At this stage, letting h = 0 will get us nowhere as we'll end up with 0 / 0. The trick is to simplfy first.
G = 2xh / h + h^2 / h = 2x + h.
Now we can safely let h become zero as we are left with 2x. So we can conclude f'(x) = 2x
Please let me know if this makes sense so far.
Bob

I see the results and I tried myself to make B a lot smaller like B = (3.0000000000001,8.0000000000001).
but it sounds wierd.

Looking at it like this :

seeing that both = goes to 0 and = goes to 0. I know what do you mean but it is so wierd for me. I heard something about ratio of "y" that is bigger than "x".

But when I used this example :

in a point "x = 1" so I had this result :

so y = 1 and y = 1.01 here it looks like the difference is small but for "x" it is 1.1-1 = 0.1 and for "y" the difference is 1.01 - 1 = 0.01 hmmm maybe this is not a good example. Because here it looks like the difference for

is going to be closer to 0 than for

Sorry for this stupid question, it sounds wierd to not understand this

Bob · 2023-01-21 20:29:32

Take care with the arithmetic.

A = ( 1 , 1 ) B = ( 1.1, 1.0201)

Bob

Jack Omar · 2023-01-22 06:57:28

Bob wrote:

Take care with the arithmetic.
A = ( 1 , 1 ) B = ( 1.1, 1.0201)
Bob

Ah yes you are right.
But this still shows that

is smaller than so basically is going faster to 0 than . It's hard to explain my point. I mean I see that this won't be 0. But seeing that these small numbers are still getting smaller and smaller it somehow give a normal not small number. Which is wierd for me.

Bob · 2023-01-22 23:52:28

A = ( 1 , 1 ) B = ( 1.01, 1.0201)

G = ( 1,0201 - 1 ) / ( 1.01 - 1 ) = 0.0201 / 0.01 = 2.01 / 1 = 2.01

Sorry I missed a zero in my previous post.

There's an old story about a frog which I'll adapt for our purposes here.

Let's imagine a frog that can make a mighty leap of one metre, but is so exhausted by the attempt that the next leap is only half a metre, and the next one quarter of a metre and so on. Each leap is half the one before. How far can the frog travel?

First leap = 1 metre
First plus second = 1.5 m
Three leaps = 1.75 m
Four leaos = 1.825 m
Five leaps = 1.9375 m
.....
I set this up on a spreadsheet using formulas to generate the values quickly. Here's the formulas I set up.
1 =1
=A1*2 =B1+1/A2
=A2*2 =B2+1/A3
=A3*2 =B3+1/A4
=A4*2 =B4+1/A5
=A5*2 =B5+1/A6
=A6*2 =B6+1/A7
=A7*2 =B7+1/A8
=A8*2 =B8+1/A9
=A9*2 =B9+1/A10
=A10*2 =B10+1/A11
=A11*2 =B11+1/A12
=A12*2 =B12+1/A13
=A13*2 =B13+1/A14
=A14*2 =B14+1/A15
=A15*2 =B15+1/A16
=A16*2 =B16+1/A17
=A17*2 =B17+1/A18
=A18*2 =B18+1/A19
=A19*2 =B19+1/A20

And here's some output:

1 1
2 1.5
4 1.75
8 1.875
16 1.9375
32 1.96875
64 1.984375
128 1.9921875
256 1.99609375
512 1.998046875
1024 1.999023438
2048 1.999511719
4096 1.999755859
8192 1.99987793
16384 1.999938965
32768 1.999969482
65536 1.999984741
131072 1.999992371
262144 1.999996185
524288 1.999998093

As you can see the total distance gets closer and closer to 2. Using MS Excel and accuracy set to 9 decimal places the values all came out as 2 after 32 rows. The 'gap' between the true answer and 2 has got so small that it's beyond the accuracy level of the software to show it.

Is it reasonable to say that eventually the frog reaches 2 metres? We would get to the point where, if you put a target tape at 2 metres, you wouldn't be able to see the frog hadn't got there.

That's what is behind the idea of a limit. We can get as close as you want to the actual gradient by getting closer and closer using B getting closer to A. In practice it works so it is widely accpeted by mathematicians as a good method.

Note: Some texts use Δx for the little bit in the x direction and some use 'h'. I think the first notation was used by Newton and the second by Leibniz. I'm happy to use either.

Here's some homework for you to try:

Work out the differentiated function by working from first principles here:

Bob

Jack Omar · 2023-01-27 07:43:53

Bob wrote:

As you can see the total distance gets closer and closer to 2. Using MS Excel and accuracy set to 9 decimal places the values all came out as 2 after 32 rows. The 'gap' between the true answer and 2 has got so small that it's beyond the accuracy level of the software to show it.
Is it reasonable to say that eventually the frog reaches 2 metres? We would get to the point where, if you put a target tape at 2 metres, you wouldn't be able to see the frog hadn't got there.
That's what is behind the idea of a limit. We can get as close as you want to the actual gradient by getting closer and closer using B getting closer to A. In practice it works so it is widely accpeted by mathematicians as a good method.
Note: Some texts use Δx for the little bit in the x direction and some use 'h'. I think the first notation was used by Newton and the second by Leibniz. I'm happy to use either.

I can see your point in the post, however isn't my limit a it different ?
I also see that the frog is in fact reaching very closely to these 2 meters, so that I can say he reached 2 meters. The only thing is that my example is a little bit different. In which for the frog there was only one limit (distance he reaches) and in my example there is "x" and "y" limit and they divide each othery "y/x" also both of them have limit to 0.
I just couldn't get it that both limits are going to 0 but I get a normal number. Both of them reaches so close to 0. But dividing them gives me a normal number. It is just unintuitive. I see that the "y" is bigger than "x" but this thought that both of them reaches 0 really messes my mind. Sorry for being so stubborn.

Bob wrote:

Here's some homework for you to try:
Work out the differentiated function by working from first principles here:

Bob

=> => => =>

I can somehow calculate it, it's just a matter of both limits which is "y" and "x" which are going to 0 and from that I get a number.

Bob · 2023-01-28 05:26:00

hi Jack,

HW all correct, well done!

So you can apply the method but don't know why it works. For the moment let's just be grateful that you can do this. I'll have another think and see if I can find another way to justify what you're doing. Watch this space.

Bob

Jack Omar · 2023-01-28 13:04:10

Bob wrote:

hi Jack,
HW all correct, well done!
So you can apply the method but don't know why it works. For the moment let's just be grateful that you can do this. I'll have another think and see if I can find another way to justify what you're doing. Watch this space.
Bob

Sure thing ! I will wait.
Of course I am not sure if what I said make any sense or if you understood it I basically understand somehow how limits works that if this delta y and delta x is going to zero then both delta y and delta x are a very very small numbers close to 0 but they are not 0. The fact is that both small numbers are divided by each other and both of them are trying to reach this "0". It is so confusing seeing also this picture :

Like why delta "y" is bigger than delta "x" what if delta "y" is smaller than delta"x" then does it mean that the number delta "y" is getting closer to 0 so it should be 0/small number. It is so wierd for me I don't know how to explain it. Tell me if what I said is wierd or not understandable, I'll try to make it more clear.

Hicies87 · 2023-02-07 07:29:08

The example you're giving is slightly different from the frog reaching a distance of 2 meters, and it's essential to understand that limits can behave differently in different scenarios. It sounds like you're struggling with the idea of both "x" and "y" limits approaching 0, but when divided, they give you an average number. This is a common phenomenon in mathematics, and it has to do with the way that limits work. Suppose you're interested in learning more about this topic. In that case, you might find it helpful to check out some resources on calculus, such as this one: https://plainmath.net/calculus-1/15773- … -important because I access it every time I don't understand any math homework questions. It's normal to struggle with new concepts at first. I didn't get a task where I had to find the function's local maximum and minimum values and saddle points. The important thing is that we're asking questions and seeking out answers. It helps me keep up the excellent work!

Last edited by Hicies87 (2023-02-17 01:31:21)

Jack Omar · 2023-02-11 09:45:18

Bob wrote:

hi Jack,
HW all correct, well done!
So you can apply the method but don't know why it works. For the moment let's just be grateful that you can do this. I'll have another think and see if I can find another way to justify what you're doing. Watch this space.
Bob

Hicies87 wrote:

Can you provide clarification on what you meant by "limits works" and why the concept is confusing to you, particularly with the use of the picture?

What did I mean by "limits works" I think it was this part : "I basically understand somehow how limits works that if this delta y and delta x is going to zero". I tried to say how do I understood the term : limits, So I understood it in a way that delta x and delta y are going to 0 but will never reach this value so it is basically a very very very small number.

When I look at the picture let's say the second one it says that if I want to find a slope of a point then it would be 0/0 but this is incorrect. So instead there are limits (lim) which are used to find that slope using a very small number. So if delta y and delta x are very small then why do they give a normal number (normal I mean not a small number like 0,00000001).
I just can't understand that delta x which will be eventually 0,0000000000000001 or smaller and delta y which is also small but somethimes delta y is smaller than delta x or delta y is bigger than delta x, so somehow they give a normal number and doesn't decrease.

So I don't understand why the number doesn't decrease when I take even smaller numbers when if they were that small like 0/0 then is would be incorrect. So logically taking even smaller number should give me smaller result but instead I have a constant number.

From the first picture it shows that the difference is really small and shrinks even further.

I don't know how to better describe my confusion. It is wierd for me that somehow I get a normal number what if delta y is smaller than delta x then it indicates that delta y is closer to 0 so getting smaller numbers shouls give me 0 but it doesn't. I'm just stuck with this imagination that when delta y and delta x are going to 0 and both of them are very small then why do I have a non 0 and a normal number.

Let's say y = x^2. So for x1 = 1 and for x2 = 1.1 then y1 = 1 and y2 = 1.21 here delta y is bigger than delta x so I get a normal number ? Even if the number gets smaller why doesn't the value change ?
Next example is : y = x/2. So for x1 = 1 and for x2 = 1.1 then y1 = 0.5 and y2 = 0.55 so now delta x is bigger than delta y and I still have a normal number ? And not 0 ? delta y is getting closer to 0 than delta x. I don't get it.

Thanks for the reply and sorry for not responding.

Last edited by Jack Omar (2023-02-11 09:56:37)

Bob · 2023-02-11 22:46:47

hi Jack,

Consider this sequence of calculations:

10/5, 4/2, 2/1, 1/0.5, 0.1/0.05, 0.000001/0.0000005 etc

The top number is getting smaller; the bottom number is getting smaller; but every division gives the answer 2.

So if we generalise this as 2x/x as x tends to zero, then what's wrong with saying it's 2.

And you can easily reverse the calculation to make the top number smaller than the bottom number:

5/10, 2/4, 1/2, 0.5/1, 0.05,0.1, 0.0000005/0.000001

This time the limit is 1/2

Bob

Math Is Fun Forum

#1 2023-01-16 07:43:28

I need help understanding fuctions.

#2 2023-01-16 22:48:19

Re: I need help understanding fuctions.

#3 2023-01-21 05:29:34

Re: I need help understanding fuctions.

#4 2023-01-21 07:24:08

Re: I need help understanding fuctions.

#5 2023-01-21 09:19:45

Re: I need help understanding fuctions.

#6 2023-01-21 20:29:32

Re: I need help understanding fuctions.

#7 2023-01-22 06:57:28

Re: I need help understanding fuctions.

#8 2023-01-22 23:52:28

Re: I need help understanding fuctions.

#9 2023-01-27 07:43:53

Re: I need help understanding fuctions.

#10 2023-01-28 05:26:00

Re: I need help understanding fuctions.

#11 2023-01-28 13:04:10

Re: I need help understanding fuctions.

#12 2023-02-07 07:29:08

Re: I need help understanding fuctions.

#13 2023-02-11 09:45:18

Re: I need help understanding fuctions.

#14 2023-02-11 22:46:47

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