Question

Suppose that f satisfies the equation f(x + y) = f(x) + f(y) + x^2y + xy^2 for all
real numbers x and y.

Answer 1

Suppose further that \[\lim_{x \rightarrow 0} f(x)/x=1\]
a) Find f(0)
b) Find f'(0)
c) Find f'(x)

Answer 2

Here is my working f(x+y)=f(0) therefore x=-y
So f(x+(-x))=f(x)+f(-y) +0 +0
f(0)=f(x)+f(-x)=0
Is this wrong?

Answer 3

It looks correct up to this point:
f(0) = f(x) + f(-x)

not sure why you're saying that =0 though.
And I'm not sure if this is the approach they wanted,
hmm still thinking about it :)

Answer 4

\[f \prime(0)=\lim_{h \rightarrow 0} f(0+h)-f(0)/h=\lim_{h \rightarrow 0} f(h)-0/h=1\]

Answer 5

Can we say that x=0

Answer 6

\[\large\rm f(\color{orangered}{x}+y)=f(\color{orangered}{x})+f(y)+\color{orangered}{x}^2y+\color{orangered}{x}y^2\]\[\large\rm f(\color{orangered}{0}+y)=f(\color{orangered}{0})+f(y)+\color{orangered}{0}^2y+\color{orangered}{0}y^2\]\[\large\rm f(y)=f(0)+f(y)\]So that's where you're getting f(0)=0, ok ok clever.

Answer 7

*Subtracting f(y) from each side*
just in case that step wasn't clear.

Answer 8

I dunno if that's right, bahhh we need someone smart

Answer 9

@ganeshie8 @SithsAndGiggles

Answer 10

@zepdrix lol, i think the work you added above is right :) but i am not sure of my approach x=-y

Answer 11

(a) and (b) look right to me. Setting \(x=-y\) yields
\[f(0)=f(x)+f(-x)\]and setting \(x=0\) yields
\[f(0)=2f(0)\implies f(0)=0\]
For the derivative at \(0\), that follows immediately from the definition of the derivative and the given behavior of \(\dfrac{f(x)}{x}\) near \(x=0\) :
\[f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\implies f'(0)=\lim_{x\to0}\frac{f(x)}{x}=1\]

Answer 12

Can i ask another question is the limit as x approaches 0 of f(x)/x=1 is that the derivative at f'(0)

Answer 13

Yes

Answer 14

Thank you @SithsAndGiggles and @zepdrix for your help :)

Answer 15

f'(x)=lim as h-->0 f(x+h)-f(x)/h
=lim as h-->0 f(x)+f(h)+x^2h+xh^2-f(x)/h
=1+lim h-->0 hx(x+h)/h=1+x^2

Answer 16

Yes, that's reasonable.