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Limits and deriv's: f satisfies f(x+y) = f(x)+f(y)+x^y+xy^2
- Thread starter kc2010
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Re: Limits and derivatives
Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
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
I have no clue where to beginSubhotosh 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.
BigGlenntheHeavy
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- Mar 8, 2009
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\(\displaystyle Given: \ f(x+y) \ = \ f(x)+f(y)+x^{2}y+xy^{2} \ and \ \lim_{x\to0}\frac{f(x)}{x} \ = \ 1, \ x \ and \ y \ are \ real \ numbers.\)
\(\displaystyle a) \ Find \ f(0). \ Let \ y \ = \ 0, \ then \ we \ have: \ f(x+0) \ = \ f(x)+f(0)+x^{2}(0)+x(0)^{2}\)
\(\displaystyle f(x) \ = \ f(x)+f(0)+0+0, \ f(0) \ = \ f(x)-f(x) \ = \ 0, \ Hence \ f(0) \ = \ 0.\)
\(\displaystyle For \ b \ and \ c, \ do \ c \ first.\)
\(\displaystyle Let \ x \ = \ h, \ then \ f(y+h) \ = \ f(h)+f(y)+h^{2}y+hy^{2} \ and \ \lim_{h\to0}\frac{f(h)}{h} \ = \ 1.\)
\(\displaystyle So, \ f(y+h)-f(y) \ = \ f(h)+h^{2}y+hy^{2}, \ \frac{f(y+h)-f(y)}{h} \ = \ \frac{f(h)+h^{2}y+hy^{2}}{h}\)
\(\displaystyle f \ ' \ (y) \ = \ \lim_{h\to0}\frac{f(y+h)-f(y)}{h} \ = \ \lim_{h\to0}\frac{f(h)+h^{2}y+hy^{2}}{h}\)
\(\displaystyle f \ ' \ (y) \ = \ \lim_{h\to0}\bigg[\frac{f(h)}{h}+hy+y^{2}\bigg] \ = \ 1+y^{2}, \ \lim_{h\to0}\frac{f(h)}{h} \ = \ 1\)
\(\displaystyle Therefore, \ f \ ' \ (y) \ = \ 1+y^{2}, \ f \ ' \ (0) \ \ = \ 1 \ (b), \ and \ f \ ' \ (x) \ \ = \ 1+x^{2} \ (c).\)
\(\displaystyle Check: \ \int[f \ ' \ (x)]dx \ = \ \int(1+x^{2})dx, \ f(x) \ = \ x+\frac{x^3}{3}+C\)
\(\displaystyle f(0) \ = \ 0 \ = \ C, \ C \ = \ 0, \ so \ f(x) \ = \ x+\frac{x^{3}}{3}, \ f \ ' \ (x) \ = \ 1+x^{2} \ and \ f \ ' \ (0) \ = \ 1. \ QED\)
\(\displaystyle a) \ Find \ f(0). \ Let \ y \ = \ 0, \ then \ we \ have: \ f(x+0) \ = \ f(x)+f(0)+x^{2}(0)+x(0)^{2}\)
\(\displaystyle f(x) \ = \ f(x)+f(0)+0+0, \ f(0) \ = \ f(x)-f(x) \ = \ 0, \ Hence \ f(0) \ = \ 0.\)
\(\displaystyle For \ b \ and \ c, \ do \ c \ first.\)
\(\displaystyle Let \ x \ = \ h, \ then \ f(y+h) \ = \ f(h)+f(y)+h^{2}y+hy^{2} \ and \ \lim_{h\to0}\frac{f(h)}{h} \ = \ 1.\)
\(\displaystyle So, \ f(y+h)-f(y) \ = \ f(h)+h^{2}y+hy^{2}, \ \frac{f(y+h)-f(y)}{h} \ = \ \frac{f(h)+h^{2}y+hy^{2}}{h}\)
\(\displaystyle f \ ' \ (y) \ = \ \lim_{h\to0}\frac{f(y+h)-f(y)}{h} \ = \ \lim_{h\to0}\frac{f(h)+h^{2}y+hy^{2}}{h}\)
\(\displaystyle f \ ' \ (y) \ = \ \lim_{h\to0}\bigg[\frac{f(h)}{h}+hy+y^{2}\bigg] \ = \ 1+y^{2}, \ \lim_{h\to0}\frac{f(h)}{h} \ = \ 1\)
\(\displaystyle Therefore, \ f \ ' \ (y) \ = \ 1+y^{2}, \ f \ ' \ (0) \ \ = \ 1 \ (b), \ and \ f \ ' \ (x) \ \ = \ 1+x^{2} \ (c).\)
\(\displaystyle Check: \ \int[f \ ' \ (x)]dx \ = \ \int(1+x^{2})dx, \ f(x) \ = \ x+\frac{x^3}{3}+C\)
\(\displaystyle f(0) \ = \ 0 \ = \ C, \ C \ = \ 0, \ so \ f(x) \ = \ x+\frac{x^{3}}{3}, \ f \ ' \ (x) \ = \ 1+x^{2} \ and \ f \ ' \ (0) \ = \ 1. \ QED\)