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geerky42

  • one year ago

I need a quick favor: I want strictly increasing elementary function \(f(x)\) such that \(f(0)=0\), \(f\left(\frac12\right)=1\), and \(\displaystyle \lim_{x\to1^-}f(x) = \infty\). I just want one example of function that satisfies given conditions. Cannot figure this out...

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  1. Loser66
    • one year ago
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    How about this \(f(x) =\dfrac{x-x^2}{1-x}\) f(0) =0 \(f(1/2) =\dfrac{1/2-1/4}{1-1/2}=1\) as x approach 1 from the left , lim --> infinitive Does it work?

  2. Loser66
    • one year ago
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    and if we simplify, we get f(x) =x, it is increasing, right?

  3. ybarrap
    • one year ago
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    You think this works? $$ f(x)=\frac{x(x+\frac{3}{2})}{x-1}\\ $$

  4. ybarrap
    • one year ago
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    http://www.wolframalpha.com/input/?i=x%28x%2B3%2F2%29%2F%28x-1%29

  5. Loser66
    • one year ago
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    @ybarrap but your graph is decreasing from x =-1

  6. ybarrap
    • one year ago
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    Or maybe... $$ f(x)=\frac{x(x+\frac{3}{2})}{1-x}\\ $$ http://www.wolframalpha.com/input/?i=x%28x%2B3%2F2%29%2F%281-x%29

  7. geerky42
    • one year ago
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    @ybarrap Not exactly what I needed, but I modified it to \(f(x) = -\dfrac{x(x+\frac32)}{2(x-1)}\), which works for me. Thanks!

  8. ybarrap
    • one year ago
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    Yep, that's it!

  9. geerky42
    • one year ago
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    @Loser66 Well, I want \(\lim_{x\to1^-}f(x)\) to be \(\infty\),

  10. geerky42
    • one year ago
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    But I got what I needed, thanks!

  11. Loser66
    • one year ago
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    ok

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