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Math Help - inverse of 0.5^2

  1. #1
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    inverse of 0.5^2

    What is the inverse function of 0.5^2?
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  2. #2
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    The question seems strange. The inverse of a number is a number, and the inverse of a function is a function (or, more generally, a relation). Here you seem to claim that the inverse of a number is a function.
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  3. #3
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    As emakarov said, in order to have an inverse function, you have to have a function! 0.5^2= 0.25 is a number, not a function. If you mean f(x)= x^2, for x non-negative, then its inverse is f^{-1}(x)= \sqrt{x}. And, of course, f^{-1}(0.25)= 0.5. As long as f has an inverse, if f(a)= b, then f^{-1}(b)= a.
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  4. #4
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    I meant to write 0.5^x. I know the inverse of for instance 2^x but not for values less than 1. Also I don't know how to find the inverse function of 2^-x which has a similar graph to 0.5^x.
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  5. #5
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    Quote Originally Posted by Stuck Man View Post
    I meant to write 0.5^x. I know the inverse of for instance 2^x but not for values less than 1. Also I don't know how to find the inverse function of 2^-x which has a similar graph to 0.5^x.
    Well (0.5)^x=2^{-x} so the inverse is -\log_2(x)
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  6. #6
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    That is not it.
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  7. #7
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    Quote Originally Posted by Stuck Man View Post
    That is not it.
    Why do you say that?

    If f(x)=(0.5)^x~\&~g(x)=-\log_2(x) then f(g(x))=g(f(x))=x.
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  8. #8
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    The graph does not look right.
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  9. #9
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    Try g(x)=-\dfrac{\ln(x)}{\ln(2)}
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  10. #10
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    I was using that since my graph program does not support multible bases. I still think it doesn't look right. There is not any image of the negative values.
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  11. #11
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    I tried it and it looks right to me... Remember that the graph of the inverse function will look like it's been reflected along the x=y line. This doesn't mean there won't be any overlap of the functions, and it doesn't mean the functions will be in opposing quadrants, which I think is what you're looking for.

    Think about it. If f(x) returns y, then g(y) returns x if g(x) and f(x) are inverse functions. In this case, where f(x) = y = 2^{(-x)}, you can have any real x you want, but y will always be positive, no matter what your x value is. So basically, the function f(x) in this case takes ANY x value, and returns a positive y value. What does that mean about its inverse function? Well... that means that its inverse function will take any positive y value, and return some real x value (no restrictions). Since you're taking the inverse function of x, that means in this case your x will always be positive and your y will span across the real numbers.

    Look at your graph again. Does it look like a reflection about the line y=x?
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  12. #12
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    Quote Originally Posted by Stuck Man View Post
    There is not any image of the negative values.
    Think about it: (0.5)^x>0 for all x.
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  13. #13
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    The logarithm function can only be used with positive values. The domain of the inverse function would be negative real values < 0.
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  14. #14
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    I was expecting a reflection on the y=-x line. I'll look at it again now. Plato, I meant negative x values.
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  15. #15
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    No. For values of x < 0, the domain will be > 0.

    EDIT: Sorry, I meant for x being a fraction.
    Last edited by Unknown008; November 10th 2010 at 06:14 AM.
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