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mathmath333

  • one year ago

find the range of the function \(f\)

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  1. mathmath333
    • one year ago
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    let \(\large \color{black}{\begin{align} f=\left\{\left(x,\ \dfrac{x^2}{1+x^2}\right):\ x\in \mathbb{R}\right\} \hspace{.33em}\\~\\ \end{align}}\) be a function from \(\mathbb{R}\) to \(\mathbb{R}\) .Determine the range of \(f\)

  2. xapproachesinfinity
    • one year ago
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    seems to me all real numbers?

  3. anonymous
    • one year ago
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    oh no

  4. anonymous
    • one year ago
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    a fraction is only 1 if the numerator is equal to the denominator

  5. welshfella
    • one year ago
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    no it cxannot have negative values

  6. anonymous
    • one year ago
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    and it is unlikely that \(x^2+1=x^2\)

  7. xapproachesinfinity
    • one year ago
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    oh miss reading the function

  8. anonymous
    • one year ago
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    also you notice that everything in sight is positive right? there is an actual way to do this

  9. anonymous
    • one year ago
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    or just wing it the least it can be is zero, since everything is positive unless x is zero and it cannot reach 1 since the numerator is smaller than the denominator

  10. anonymous
    • one year ago
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    making the range \[[0,1)\] otherwise you have to set \[y=\frac{x^2}{1+x^2}\] and solve for \(x\) which is a colossal pain

  11. mathmath333
    • one year ago
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    i was looking for a way which would be different from just observation

  12. anonymous
    • one year ago
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    there is nothing wrong with thinking, not matter what you math teacher tells you

  13. anonymous
    • one year ago
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    but yes, there is a way solve \[y=\frac{x^2}{1+x^2}\] for \(x\)

  14. anonymous
    • one year ago
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    you will get a quadratic equation in \(y\) you then take the discriminant and make sure it is greater than or equal to zero enjoy, it is a pain in the neck

  15. mathmath333
    • one year ago
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    i mean \(y(y-1)\geq 0\)

  16. mathmath333
    • one year ago
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    is that correct

  17. xapproachesinfinity
    • one year ago
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    is it not y(1-y)>=0

  18. anonymous
    • one year ago
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    looks right

  19. mathmath333
    • one year ago
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    ya this one ^

  20. anonymous
    • one year ago
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    i think if you solve for \(x\)you get \[x=\sqrt{\frac{y}{1-y}}\]

  21. anonymous
    • one year ago
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    domain of that one is \[[0,1)\] so range of your function is the same

  22. anonymous
    • one year ago
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    really i think this is a butt load more work, than necessary common sense often works in math too

  23. mathmath333
    • one year ago
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    but why this wont work \(y(1-y)>=0 \)

  24. anonymous
    • one year ago
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    why strictly greater than?

  25. anonymous
    • one year ago
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    oh i see you have \(\geq\)

  26. mathmath333
    • one year ago
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    it will be \(1\geq y\geq 0\)

  27. anonymous
    • one year ago
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    but that is not right, because \(\geq\) includes 1

  28. anonymous
    • one year ago
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    and there is no way for your function to be 1

  29. mathmath333
    • one year ago
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    thats why was asking why that method backstabbed

  30. anonymous
    • one year ago
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    whereas solving \[\frac{y}{1-y}\geq 0\] does not include 1

  31. mathmath333
    • one year ago
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    ok thnx

  32. anonymous
    • one year ago
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    ok i see my mistake i mislead you

  33. anonymous
    • one year ago
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    you do not get a quadratic, you solve \[y=\frac{x^2}{1+x^2}\] \[yx^2+y=x^2\\ yx^2-x^2=-y\\ (y-1)x^2=-y\\ x^2=\frac{y}{1-y}\\ x=\sqrt{\frac{y}{1-y}}\]

  34. anonymous
    • one year ago
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    then solve \[\frac{y}{1-y}\geq0\] you get \[[0,1)\]

  35. anonymous
    • one year ago
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    sorry about that i could never win against a king's gambit either, not matter whether i accepted or declined

  36. mathmath333
    • one year ago
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    haha

  37. mathmath333
    • one year ago
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    i have good record against kings gambit by black

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