2007-07-27 08:51:34 · 8 answers · asked by no idea 2 in Science & Mathematics ➔ Other - Science
NO.
If they are the "same" then they are =
so
f(x) = g(x)
or
x - 1 = x^2 - 1/x +1
multiply all elements by x
x^2 - x = x^3 -1 + x
substituting 0 for x
0 - 0 = 0 - 1 + 0
0 does not equal -1
2007-07-27 09:01:05 · answer #1 · answered by vpi61 2 · 0⤊ 0⤋
The second fuction can be written as
(x-1)(x+1)/x+1 because x^2-1^2=(x-1)(x+1)
you eliminate the two x+1 and what remains to you is
x-1
that means that g(x)=x-1 and so as f(x)=x-1 they are the same
I'm telling you they ARE the same. Don't trust those others. I'll prove you that I'm right.
Let's suppose that x=2 ok? Than
f(2)=2-1=1
g(2)=2^2-1/2+1=4-1/3=3/3=1
You see? You can try with any value of x and you'll see that f and g are the same function. Trust me. I'm the only one to have an A in Maths. :-)
2007-07-27 21:04:06 · answer #2 · answered by Tasha 3 · 0⤊ 1⤋
No they are not the same despite Tasha's claim. GTB had it in his or her hands but did not draw the right conclusion either.
They are the same if and only if the range and domain of f(x) is equal to the range and domain of g(x) and vice versa. Another way to get at this is to ask "Is there a value of x for which f(x) is not equal to g(x)?"
The answer is, as GTB pointed out x = -1. f(-1) = -2 but g(-1) is not defined and hence f(-1) does not equal g(-1). So f(x) is not the same as g(x).
Having said that, there are the same every where else.
HTH
Charles
2007-07-27 22:37:22 · answer #3 · answered by Charles 6 · 0⤊ 0⤋
yes. If you break up the top part of the 2nd function (x^2-1) you will get (x+1)(x-1)/(x+1) because of the Difference of Squares rule. the (x+1) terms cancel out on top and bottom and you are left with (x-1)
2007-07-27 15:54:44 · answer #4 · answered by Anonymous · 0⤊ 0⤋
g(x) = x^2-1/x+1 = (x+1)(x-1)/(x+1) = (x-1)
looks like f and g are the same
technically you can only divide by x+1 when x not equal to -1 because you can't divide by zero
2007-07-27 15:55:49 · answer #5 · answered by GTB 7 · 0⤊ 0⤋
No. When you take the derivative of each individual function, they do not yield the same answer. Especially obvious when you take the derivative and then try to plug values in for "x".
2007-07-27 15:56:07 · answer #6 · answered by Curious 1 · 0⤊ 0⤋
No, there is nothing the same about them. They have different domains, and they only equal eachother at x=1.
That wasn't a take home test problem was it?
2007-07-27 15:58:37 · answer #7 · answered by Jay Linne 2 · 0⤊ 0⤋
plug in numbers and graph them
2007-07-27 15:54:19 · answer #8 · answered by PhantomRN 6 · 0⤊ 0⤋