the graph of r(x) has a slant asymptote of the form
y = 2x + 4 ??? is this correct or is it y = 2x + 2 ???
2006-09-03 09:44:28 · 6 answers · asked by Doug 2 in Education & Reference ➔ Higher Education (University +)
g(x) = x^2-x+1 = 0 has no real roots so you're in lucky f(x)/g(x) will be everywhere differentiable since g(x) <> 0 at any x
You could use the product rule for f(x),[1/g(x)] if you forgot the quotient rule but ...
The quotient rule is
d/dx[f(x)/g(x)] = (g(x)f'(x) - f(x)g'(x)) / [g(x)]^2
g(x) = x^2-x+1
g'(x) = 2x-1
f(x) = 2x^3-5x^2-1
f'(x) = 6x^2-10x
g(x)f'(x) = 6x^4-16x^3+16x^2-10x
f(x)g'(x)= 4x^4-12x^3+5x^2-2x+1
g(x)f'(x)-f(x)g'(x) = 2x^4-4x^3+11x^2-8x-1
g(x)^2 = x^4-2x^3+3x^2-2x+1
So r'(x) =
(2x^4-4x^3+11x^2-8x-1)
------------------------------
(x^4-2x^3+3x^2-2x+1)
lim x->infinity r'(x) = 2
2006-09-03 10:41:38 · answer #1 · answered by Andy S 6 · 1⤊ 0⤋
=The answer is in your Calculus book
2006-09-03 09:45:50 · answer #2 · answered by Anonymous · 0⤊ 0⤋
uh confusing!
2006-09-03 09:49:41 · answer #3 · answered by manateeluver32 3 · 0⤊ 0⤋
whoa
all i gotta say is
GOOD LUCK!
2006-09-03 09:46:53 · answer #4 · answered by Anonymous · 0⤊ 0⤋
OKKKKKKKKKKK?I'M GUESSING YOUR IN COLLAGE!
2006-09-03 09:47:01 · answer #5 · answered by Anonymous · 0⤊ 0⤋
is this einstins home work??
2006-09-03 09:45:49 · answer #6 · answered by chris 3 · 0⤊ 0⤋