Solving for f(x) in a Function with Given Equations and Variables

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Homework Help Overview

The discussion revolves around a function f that satisfies two given equations, one involving a functional equation and the other a limit condition. Participants are tasked with finding specific values and derivatives of the function at certain points.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the functional equation and the limit condition, questioning how to derive values for f(0) and f'(0). Some suggest substituting specific values into the equations to gain insights, while others discuss the definition of the derivative and its application in this context.

Discussion Status

The conversation is ongoing, with various interpretations and approaches being explored. Some participants have offered insights into the calculations and reasoning behind the derivatives, while others express confusion and seek clarification on the assumptions being made.

Contextual Notes

There is a noted lack of clarity regarding the definitions and the setup of the problem, with participants questioning the validity of certain steps and the implications of dividing by zero in their calculations.

ARYT
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functions ?

Homework Statement



I apologize for the title, I don't know what to call these kind of problems.

Q3. Suppose f is function that satisfies the equation

equation 1

For all numbers x and y, suppose also that

equation 2

Find: (i). f(0), (ii). f’(0), (iii). f’(x)



Homework Equations



?

The Attempt at a Solution



I have no idea. So what?
 

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hi Aryt

pictures take a while to get viewable... so can't see yet, can you enter the problem otherwise?
 


equation 1: f(x+y)=f(x)+f(y)+x^2 y+xy^2

equation 2: lim f(x)/x =1 while x tends to zero
 


is that all the info...? if so, then how about considering f(1+0)?

then think about a derivative from first limits at zero...

anyway, see what you think ;)
 


I think we should write sth like: f(-1+1) = f(0) = f(1)- f(-1)+(1)^2 (-1) + (1)(-1^2)
which will be f(1)-f(-1). and we know that when x tends to 0, f(x)/x = 1
so f(1)/1 = 1 and also f(-1)/-1 = 1
Therefore f(0) = 1-1=0 or I dunno, maybe 1+1=2 .

Is it a possible answer for (i)?

ahhhhhh, what kind of question is this? :(
 


I think that it is rather a good question.
Get to the definition of f'(x).

f'(x)= [f(x+h) - f(h)}/h . Where h tends towards zero. Now think what can you do to f(x+h). Once done with that. You wil get to a dead end. But you can get rid of that by knowing the value of f(0). The value of f(0) cab be found out by replacing x and y in the original equation by 0 and 0. Once done with that you will get the value of F(0).
 


well, if we go for 0 and 0, we will have: f(0) = f(0)+f(0) that is f(0)=2f(0)
and 1=2 ! Great!
 


Well its obvious isn't it. You are divinding by zero. Hence you get ridiculous statements like 1=2. So f(0)=0
 


so f'(0) is also zero?

and f'(x)=1?

I think I got it. Thanks
 
  • #10


i think you're there, but here's what I meant

f(x+y)=f(x)+f(y)+x^2 y+xy^2
f(0+1)=f(0)+f(1)+0^2.1+0.1^2
f(1)=f(0)+f(1)
0=f(0)
 
  • #11


ARYT said:
so f'(0) is also zero?

and f'(x)=1?

I think I got it. Thanks

I haven't calculated the value of f'(x). Even if it is equal to 1 how can the value of f'(0) be equal to zero considering that f'(x)=1
 
  • #12


yeah sorry i misread your last post

how did you get f'(0) = 0, is that a guess?
calculate f'(0) first as Fedex implies (use the limit information and the first principles defintino of a derivative)
 
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  • #13


f'(x)= [f(x+h) - f(h)]/h while h tends to zero.

so f'(0)=[f(0+0)-f(0)]/0
f'(0)= 0/0

?
 
  • #14


remember its not h=0, its the limit as h goes to zero
 
  • #15


We're under heavy fire C.O., Clarification is needed. :D

lol

Isn't it a bit stupid to say:
We have: f(0) = 0 and Equation 1:

we have lim [f(0+h)-f(h)]/h while h maps to 0.
Thus (h tends to 0): lim f(h)/h - lim f(h)/h = 1-1 = 0
 
  • #16


I think I've found the problem. You've cited the equation wrongly. It should be:

[f(x+h)-f(x)]/h while h tends to 0. According to equation 1, we have:

[f(x)+f(h)+x^2 h + xh^2 - f(x)] / h => lim f(h)/h + lim (x^2+xh) while h maps to zero.

So f'(x)=1+x^2

and accordign to this we have: f'(0)=1+0=1

Right?
 
  • #17


ARYT said:
f'(x)= [f(x+h) - f(h)]/h while h tends to zero.

so f'(0)=[f(0+0)-f(0)]/0
f'(0)= 0/0

?

Buffon
 
  • #18


ARYT said:
I think I've found the problem. You've cited the equation wrongly. It should be:

[f(x+h)-f(x)]/h while h tends to 0. According to equation 1, we have:

[f(x)+f(h)+x^2 h + xh^2 - f(x)] / h => lim f(h)/h + lim (x^2+xh) while h maps to zero.

So f'(x)=1+x^2

and accordign to this we have: f'(0)=1+0=1

Right?

Genuis

PS Sorry for "buffon" I just wanted to bring a dramatic end.
 
  • #19


No, It's OK. Thanks for the tips. :D

A better dramatic end: lol

Although they're killing my innovation here. For one assignment (C++ programming), I've used only one line to compile a rather complicated program without sth useless (i.e. End of File function: eof.). And the tutor told me very directly, “why are you trying to come up with your own ways”? I gave you a kind of lecture for this question. When you can't solve sth in my way, why you do it in your way?
ahhhh, it's innovation, if you add only one space to the input file, your program will fail, but mine will remain ok, I said! :)
 
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