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math problem! (finding the inverse of a function)
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by silver_deeds
on 2006-12-17 12:31:51
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Ok, I have a calculus question. My teacher isn't good at explaining things and it didn't help that the bell rang on me, so I hope the smart people here could maybe help me understand. (and I don't know how to type all of those mathematical symbols with a keyboard so please bare with me ^-^;;) It's about inverting functions and finding/showing a function's inverse. To do this you prove that f(g(x))=x as well as g(f(x))=x and therefor you find f^-1(x). Graphically, you show it by reflecting the functions through the y=x line. Of course the question arose of how do you know if the given function even has an inverse. I understand how it would be shown on a graph and solving it comes easy to me, But the proof itself I had a question about: A function does not have an inverse if f(x), substituting with a and b for each side, do not equal each other (in other words f(a)=f(b)) Examples: f(x)=x^1/2 a^1/2=b^1/2 (a^1/2)^2=(b^1/2)^2 a=b this is a 1-1 function (it has an inverse)where f^-1(x)=x^2 & x >and= 0 But: f(x)=x^2 a^2=b^2 (a^2)^1/2=(b^2)^1/2 a doesn't = +-b (the + is supposed to be above the - but I don't know how to do that or the slash through the = sign...) Finally, here is my question! Why is it that only b gets the +- when nothing different happened to a? I know that in the end, when solving for only 1 variable you solve it to be only + but I don't see how it can happen with this proof. Please explain how it would work. But, I do understand that x^2 doesn't have an inverse as well as any even root since a number in the domain can't have more than 1 number in the range, for example both 2^2 and-2^2 = 4 and that can't happen when finding the inverse. Graphing it shows this as well. I understand the concept it's the whole thing on how the +- can only appear on 1 side. Thanks for reading and any explanation would be nice! Once again, I'm sorry I don't know how to make the symbols with my keyboard and I hope it didn't confuse anybody... XD Also, could someone tell me how this would be used in real life? It would be nice to understand how this can apply fo my futur other then if I wanted to be a math teacher =P |
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Re: math problem! (finding the inverse of a function)
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The 1-1 definition says that: "A function f is one-to-one if, for every a and b in its domain, f(a) = f(b) implies a = b." (Source) That means, to find out if a function is 1-1, you see if f(a) and f(b) imply that a = b. So if f(a) = 3 and f(b) = 3, the function is 1-1 if a = b. In your second example, since a could either be +b or -b, it is not 1-1. To say that again, if f(a) = f(b) but a != b at every value, there is no inverse function. if f(a) = f(b) always means a = b, there is an inverse function. (btw, for "not equals," programmers use "!=", it's rather convenient)
>,>; Did I just say that...?
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Re: math problem! (finding the inverse of a function)
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By convention, only the top half of the square root is taken for a graph. Maybe i should explain this better. Normally: if y^2 = x then - y = +/- sqrt(x) However, by convention, when graphing this, we only take the positive. That is - if y^2 = x then y = +sqrt(x) This shows on the graph also. This is the only way to qualify a square root as a function, for if it were not this way, an input value for 'x' would have two output values. P.S. The easiest way to find inverses is to replace every y value with x and every x value with y. For example if y = 2x + 5 then the inverse is x = 2y + 5 Thus, we write it was y = (x - 5) / 2 To check, we can see if the new line/curve is symmetrical to the old one through the line y=x  |
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Re: math problem! (finding the inverse of a function)
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by silver_deeds
on 2007-01-19 15:21:34
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| Thanks alot! I'm doing a lot better in math now, too. I think... ^-^;; |