Limit Question: Finding f(0), f'(0), and f'(x) | Countless Hours of Frustration

[clarence]

Homework Statement

f(x+y)= f(x) + f(y) + y*x^2 +x*y^2 Given: lim of f(x)/x where x approaches 0 is 1
Find : 1) f(0) 2) f ' (0) 3) f ' (x)

Homework Equations

The Attempt at a Solution

countless hours with other coursemates that lead to nothing but a looping headache.

 

[statdad]

Note that

<br /> f(x+y) = f(x) + f(y) + x^2 y + x y^2<br />

means that (using x = y = 0

<br /> f(0) = f(0) + f(0) + 0<br />

What does this tell you about f(0)? Also remember the definition:

<br /> f&#039;(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{(0+h) - 0} = \lim_{h \to 0} \frac{f(h)-f(0)} h<br />

This second point relates to one thing you are given

 

[clarence]

Yep, that be true. It's the 3rd part of the question that i can't still solve. I've posted the first two parts of the question as a guide to solving the third part. So any ideas on the 3rd part? Thanks for making the second part ans more organize. My solution was a little more messy.

 

[clarence]

Note that

<br /> f(x+y) = f(x) + f(y) + x^2 y + x y^2<br />

means that (using x = y = 0

<br /> f(0) = f(0) + f(0) + 0<br />

What does this tell you about f(0)? Also remember the definition:

<br /> f&#039;(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{(0+h) - 0} = \lim_{h \to 0} \frac{f(h)-f(0)} h<br />

This second point relates to one thing you are given

but x+y=0, x can be -y. that is, x and y may not equal to zero?
so f(0)=f(x)+f(-x)?

 

[gabbagabbahey]

f&#039;(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{(x+h)-x}

But given that f(x+y)=f(x)+f(y)+x^2y+xy^2 , what is f(x+h) ?

 

[clarence]

Wow, true, I didn't see it that way.Thanks a lot. LOL