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HI I am given a function g(x)= x^2 - 1, where x takes any real value.
I am asked to show that this function is not a one-one function.
Apart from the method of horizontal line test(graphical method of drawing horizontal line and show that cuts graph more than once)
what other method can use to show its not a one-one function?
thanks! (:
g(x)=x[sup]2[/sup]-1
is defined for all R-->R
To show that g(x) is not one-one we need to show that g(x[sub]1[/sub]) = g(x[sub]2[/sub]) for some x[sub]1[/sub] ≠ x[sub]2[/sub]
Let g(x[sub]1[/sub]) = g(x[sub]2[/sub])
x[sub]1[/sub][sup]2[/sup] - 1 = x[sub]2[/sub][sup]2[/sup] - 1
x[sub]1[/sub][sup]2[/sup] = x[sub]2[/sub][sup]2[/sup]
x[sub]1[/sub][sup]2[/sup] - x[sub]2[/sub][sup]2[/sup] = 0
(x[sub]1[/sub] - x[sub]2[/sub])(x[sub]1[/sub] + x[sub]2[/sub]) = 0
x[sub]1[/sub] - x[sub]2[/sub] = 0 or x[sub]1[/sub] + x[sub]2[/sub] = 0
x[sub]1[/sub] = x[sub]2[/sub] or x[sub]1[/sub] = - x[sub]2[/sub]
i. e. x[sub]1[/sub] = x[sub]2[/sub] or x[sub]1[/sub] ≠ x[sub]2[/sub]
However.. this may not be the only proof.
Last edited by ZHero (2010-05-11 02:39:01)
If two or more thoughts intersect, there has to be a point!
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Hi estelle j;
A function for which every element of the range of the function corresponds to exactly one element of the domain. One-to-one is often written 1-1.
This function is so simple you can do it like this.
To prove that it is not 1 - 1 you just need 1 counterexample. In this case (-4)^2 - 1 = 15 and 4^2 - 1 = 15. 2 x's yielding the same y. Your knowledge that a negative times a negative = a positive is enough to find a counterexample.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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The only comment I have is that it should have been "an interesting comment" in the hidden text...
Honestly, I have no idea what it is you're looking for.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Hi Rick;
I need to fill in here. What I meant was If it passed by you and her then that was sort of like peer review. Except, in this case we are not peers. The old guy is lagging 2 miles behind and out of breath.
I have been meaning to tell you this:
Recently, a question about whether a function was defined or not came up. This function had a simple pole, I think. Although you and mathsy had already explained it to me, I still made a mistake. Jane protested but I just waved her off. Armed with some new info about removable singularities, I calmy continued. Finally, a guy, called "Dr. Rick," (Ask Dr. Math) pointed it out so bluntly that it actually sunk in.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Ah, I see. Your post was spot on, bobby.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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