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anonymous

  • 5 years ago

Let , x^2+2x, x is > or = to -1. Find its inverse and state its domain and range using interval notation. If required, use fractions, NOT decimals. I did quadratic formula and got: My answer: (-2+-sqrt(4+4x))/2 Domain of : (-oo,-1] but this answer is wrong, what did I do wrong?

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  1. anonymous
    • 5 years ago
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    I put [-1,oo) for range and that was correct...

  2. anonymous
    • 5 years ago
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    Yeah, the range is \([-1,\infty)\).

  3. anonymous
    • 5 years ago
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    \[\left( -2\pm \sqrt(2^2-4(1)(-1)\right)/2\] This is my answer if thats clearer

  4. anonymous
    • 5 years ago
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    hello anwar! mistake is the \[\pm\] this a function so only one output for each input.

  5. anonymous
    • 5 years ago
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    To find the inverse, set \(x=y^2+2y\), and try to solve for y. You will have \(y^2+2y-x=0\), by using the quadratic formula, we have: \[y={-2\pm \sqrt{4+4x} \over 2}=-1\pm \sqrt{1+x}\]. But since the range is \((-1,\infty)\); we will take only \(+\sqrt{1+x}\).

  6. anonymous
    • 5 years ago
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    Hello satellite!

  7. anonymous
    • 5 years ago
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    hello. nice latex.

  8. anonymous
    • 5 years ago
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    So, the inverse is \(\sqrt{1+x}-1\). Try to find the domain. And thank ssatellite :)

  9. anonymous
    • 5 years ago
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    ok. S owhy would the answer for the domain not be (-oo,-1]?

  10. anonymous
    • 5 years ago
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    The domain of the inverse has to be the same as the range of the original function. So, what do you think is the range of the original function, in the given domain \((-1,\infty)\)?

  11. anonymous
    • 5 years ago
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    \([-1,\infty)\)*

  12. anonymous
    • 5 years ago
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    why is \[\pm\] a mistake?

  13. anonymous
    • 5 years ago
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    A function can't have a ±, because it would fail the vertical line test then.

  14. anonymous
    • 5 years ago
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    The domain of the inverse, by the way, is \([-1,\infty)\).

  15. anonymous
    • 5 years ago
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    oh, but I thought I'm to use quadratic formula to find the inverse of f(x)

  16. anonymous
    • 5 years ago
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    in this situation

  17. anonymous
    • 5 years ago
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    let me butt in for a second. think about \[f(x)=x^2\] if you write \[x=y^2\] and solve for y you get \[y=\pm\sqrt{x}\] but this is not a function. and of course \[f(x)=x^2\] is not a one to one function. but if you restrict the domain of \[f(x)\] to x>0, then the inverse has to have range y > 0 and so you just write \[f^{-1}(x)=\sqrt{x}\] a perfectly good function.

  18. anonymous
    • 5 years ago
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    I understand that there is only one output for one input, so how am i supposed to do that?

  19. anonymous
    • 5 years ago
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    But, in the original function the domain is \([-1,\infty)\), and therefore the range of the inverse has to be \([-1,\infty)\). But if we take \(-\sqrt{x+1}\), that would lead to values that are not part of the range.

  20. anonymous
    • 5 years ago
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    how do I express that?

  21. anonymous
    • 5 years ago
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    It's already expressed :D

  22. anonymous
    • 5 years ago
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    so, my question i guess is what is the range? [-1,oo)?

  23. anonymous
    • 5 years ago
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    Yep.

  24. anonymous
    • 5 years ago
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    Im confused. If it leads to wrong values, then why does it work?

  25. anonymous
    • 5 years ago
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    Im confused. If it leads to wrong values, then why does it work?

  26. anonymous
    • 5 years ago
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    That would be only If we take the minus the square as part of the solution, but we didn't. So, we're in the safe side.

  27. anonymous
    • 5 years ago
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    OOOOO! I gotcha, thank you :)

  28. anonymous
    • 5 years ago
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    can I ask you different question, about absolute value? x=|y+3|

  29. anonymous
    • 5 years ago
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    trying to figure out inverse of that?

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