Let f(x) = x/(x - 1) g(x) = (x - 1)/x h(x) = 1/(x - 1)
Then (f + g)(x) =
I got ... 1/x ?
2006-09-02 11:17:16 · 8 answers · asked by Olivia 4 in Science & Mathematics ➔ Engineering
x1
2006-09-02 11:22:06 · answer #1 · answered by shacadia 2 · 0⤊ 2⤋
I multiplied f by x/x and g by (x - 1)/(x - 1) so that they'd both have a denominator of x(x - 1). Then I added f and g and I got (2x^2 - 2x + 1) / (x^2 - x). The numerator has no real roots so it can't be factored, meaning there is no simpler form for this fraction.
2006-09-02 11:23:58 · answer #2 · answered by DavidK93 7 · 0⤊ 0⤋
i think it's X^2-1/x
2006-09-02 11:24:25 · answer #3 · answered by cosmic_convoy 3 · 0⤊ 0⤋
(f + g)(x) = f(x)+g(x) = x/(x-1)+(x-1)/x
create denominator common
= x^2/[x*(x-1)] + (x-1)^2/[x*(x-1)]
add numerators
= (2x^2-2x+1)/(x^2-x)
h(x) is extraneous to this particular problem
2006-09-02 11:33:33 · answer #4 · answered by Andy S 6 · 0⤊ 0⤋
I think the answer is 0
2006-09-02 11:19:51 · answer #5 · answered by gmr1723 1 · 0⤊ 0⤋
I get
(2x^2 - 2x +1)/ (x^2-x)
(why is "h"specified?)
2006-09-02 11:26:13 · answer #6 · answered by Vincent G 7 · 0⤊ 0⤋
NOPE I SUCK AT MATH 2
2006-09-02 11:41:43 · answer #7 · answered by Tan 1 · 0⤊ 0⤋
pi
2006-09-02 11:22:57 · answer #8 · answered by Anonymous · 0⤊ 0⤋