Limits and deriv's: f satisfies f(x+y) = f(x)+f(y)+x^y+xy^2

kc2010

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Sep 23, 2009
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Suppose f is a function that satisfies the equation
f(x + y) = f(x) + f(y) +x^2y + xy^2
for all real numbers x and y. Suppose also that lim x>0 f(x)/x =1
a) find f(0)
b) find f'(0)
c) find f'(x)
 
Re: Limits and derivatives

kc2010 said:
Suppose f is a function that satisfies the equation
f(x + y) = f(x) + f(y) +x^2y + xy^2
for all real numbers x and y. Suppose also that lim x>0 f(x)/x =1
a) find f(0)
b) find f'(0)
c) find f'(x)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
Re: Limits and derivatives

Subhotosh Khan said:
kc2010 said:
Suppose f is a function that satisfies the equation
f(x + y) = f(x) + f(y) +x^2y + xy^2
for all real numbers x and y. Suppose also that lim x>0 f(x)/x =1
a) find f(0)
b) find f'(0)
c) find f'(x)

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
I have no clue where to begin
 
Given: f(x+y) = f(x)+f(y)+x2y+xy2 and limx0f(x)x = 1, x and y are real numbers.\displaystyle Given: \ f(x+y) \ = \ f(x)+f(y)+x^{2}y+xy^{2} \ and \ \lim_{x\to0}\frac{f(x)}{x} \ = \ 1, \ x \ and \ y \ are \ real \ numbers.

a) Find f(0). Let y = 0, then we have: f(x+0) = f(x)+f(0)+x2(0)+x(0)2\displaystyle a) \ Find \ f(0). \ Let \ y \ = \ 0, \ then \ we \ have: \ f(x+0) \ = \ f(x)+f(0)+x^{2}(0)+x(0)^{2}

f(x) = f(x)+f(0)+0+0, f(0) = f(x)f(x) = 0, Hence f(0) = 0.\displaystyle f(x) \ = \ f(x)+f(0)+0+0, \ f(0) \ = \ f(x)-f(x) \ = \ 0, \ Hence \ f(0) \ = \ 0.

For b and c, do c first.\displaystyle For \ b \ and \ c, \ do \ c \ first.

Let x = h, then f(y+h) = f(h)+f(y)+h2y+hy2 and limh0f(h)h = 1.\displaystyle Let \ x \ = \ h, \ then \ f(y+h) \ = \ f(h)+f(y)+h^{2}y+hy^{2} \ and \ \lim_{h\to0}\frac{f(h)}{h} \ = \ 1.

So, f(y+h)f(y) = f(h)+h2y+hy2, f(y+h)f(y)h = f(h)+h2y+hy2h\displaystyle So, \ f(y+h)-f(y) \ = \ f(h)+h^{2}y+hy^{2}, \ \frac{f(y+h)-f(y)}{h} \ = \ \frac{f(h)+h^{2}y+hy^{2}}{h}

f  (y) = limh0f(y+h)f(y)h = limh0f(h)+h2y+hy2h\displaystyle f \ ' \ (y) \ = \ \lim_{h\to0}\frac{f(y+h)-f(y)}{h} \ = \ \lim_{h\to0}\frac{f(h)+h^{2}y+hy^{2}}{h}

f  (y) = limh0[f(h)h+hy+y2] = 1+y2, limh0f(h)h = 1\displaystyle f \ ' \ (y) \ = \ \lim_{h\to0}\bigg[\frac{f(h)}{h}+hy+y^{2}\bigg] \ = \ 1+y^{2}, \ \lim_{h\to0}\frac{f(h)}{h} \ = \ 1

Therefore, f  (y) = 1+y2, f  (0)  = 1 (b), and f  (x)  = 1+x2 (c).\displaystyle Therefore, \ f \ ' \ (y) \ = \ 1+y^{2}, \ f \ ' \ (0) \ \ = \ 1 \ (b), \ and \ f \ ' \ (x) \ \ = \ 1+x^{2} \ (c).

Check: [f  (x)]dx = (1+x2)dx, f(x) = x+x33+C\displaystyle Check: \ \int[f \ ' \ (x)]dx \ = \ \int(1+x^{2})dx, \ f(x) \ = \ x+\frac{x^3}{3}+C

f(0) = 0 = C, C = 0, so f(x) = x+x33, f  (x) = 1+x2 and f  (0) = 1. QED\displaystyle f(0) \ = \ 0 \ = \ C, \ C \ = \ 0, \ so \ f(x) \ = \ x+\frac{x^{3}}{3}, \ f \ ' \ (x) \ = \ 1+x^{2} \ and \ f \ ' \ (0) \ = \ 1. \ QED
 
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