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anonymous

  • 5 years ago

f(x)=-16x^2 What is the inverse function of.. I need the answer plzzzzzz, assume that all variables are positive. Again, does anyone know the answer? Someone out there is smart, i know there is.......

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  1. amistre64
    • 5 years ago
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    x = -(16y^2) x/16 = -(y^2) sqrt(x)/4 = -y - sqrt(x)/4 = y

  2. amistre64
    • 5 years ago
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    maybe

  3. anonymous
    • 5 years ago
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    but that would come out to having an i in the final answer correct?

  4. amistre64
    • 5 years ago
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    nah, i made sure that I took out the -1 first to avoid that ...maybe lol let me look at it after drawing it

  5. anonymous
    • 5 years ago
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    amistre?

  6. amistre64
    • 5 years ago
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    yes?

  7. anonymous
    • 5 years ago
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    maybe i am tired but i think you missed this one.

  8. amistre64
    • 5 years ago
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    oh i miss alot of them ; not just this one lol

  9. anonymous
    • 5 years ago
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    ya, this thing is like impossible, it's a question on my test, what should i write down as the answer?

  10. amistre64
    • 5 years ago
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    its x = -y^2 but i dont think we can make a function out of it can we?

  11. anonymous
    • 5 years ago
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    f says square, then multiply by 16. f inverse says divide by -16, take the square root. should be \[f^{-1}(x)=\sqrt{-\frac{x}{16}}\]

  12. amistre64
    • 5 years ago
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    we can stick an absolute value in there cant we :)

  13. anonymous
    • 5 years ago
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    one to one because they restricted \[x>0\]

  14. amistre64
    • 5 years ago
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    \[y = \frac{\sqrt{|-x|}}{4}\]

  15. anonymous
    • 5 years ago
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    so function is defined only on positive x, and therefore is one to one. range is obviously negative numbers.

  16. anonymous
    • 5 years ago
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    i got f^-1=1/4(x) but i know that's wrong......

  17. anonymous
    • 5 years ago
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    no range is negative so domain of inverse is negative numbers

  18. amistre64
    • 5 years ago
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    \[\frac{\sqrt{|x|}}{4} ; D:(-\infty,0]\]

  19. amistre64
    • 5 years ago
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    maybe? lol

  20. anonymous
    • 5 years ago
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    i dont have to graph it

  21. anonymous
    • 5 years ago
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    absolute value not necessary just put \[\sqrt{\frac{-x}{4}}\]

  22. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%28 |x|%29%2F4

  23. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=sqrt%28%7Cx%7C%29%2F4

  24. anonymous
    • 5 years ago
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    ah but that is not the inverse of this function.

  25. amistre64
    • 5 years ago
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    gotta run...ciao :)

  26. anonymous
    • 5 years ago
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    this function was restricted to positive x.

  27. anonymous
    • 5 years ago
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    ok later.

  28. amistre64
    • 5 years ago
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    it is if we rectric the domain lol; like trig inverse functions

  29. anonymous
    • 5 years ago
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    more puzzles please.

  30. anonymous
    • 5 years ago
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    123456789 are you confused yet?

  31. anonymous
    • 5 years ago
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    very......i'm a high school freshman in algebra 2 honors.... this is like crazy college crap lol

  32. anonymous
    • 5 years ago
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    let me try to xplain. do you know what the graph of \[y=-x^2\] looks like?

  33. anonymous
    • 5 years ago
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    yeah, it's a prarabola pointing down

  34. anonymous
    • 5 years ago
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    but we arent doing parabloas

  35. anonymous
    • 5 years ago
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    well the point is that since it is a parabola it is not a one to one function so it doesn't have an inverse. but they made it a one to one function by restricting the domain to positive numbers. so your actual graph is not a parabola, it is just the right hand side of the parabola. hterefore it is one to one

  36. anonymous
    • 5 years ago
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    so the way you do this is write \[y=-16x^2\] replace y by x and x by y to get \[x=-16y^2\]

  37. anonymous
    • 5 years ago
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    ok, now what?

  38. anonymous
    • 5 years ago
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    then solve for y: \[x=-16y^x\] \[\frac{-x}{16}=y^2\] \[y=\frac{\sqrt{-x}}{16}\]

  39. anonymous
    • 5 years ago
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    again the typesetting is bad. the square root should be over the whole thing.

  40. anonymous
    • 5 years ago
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    so it should be \[f^{-1}(x)=\frac{\sqrt{-x}}{4}\] since the square root of 16 is 4

  41. anonymous
    • 5 years ago
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    and you don't have to worry about \[\sqrt{-x}\] being a problem because the range of your original function was negative numbers, so your domain of the inverse is negative numbers.

  42. anonymous
    • 5 years ago
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    take out the i or leave it in?

  43. anonymous
    • 5 years ago
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    x is negative so -x is positive so you can take its square root.

  44. anonymous
    • 5 years ago
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    thanks a bunch

  45. anonymous
    • 5 years ago
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    leave it in. the domain of your original function was positive numbers so the range of your inverse must be positive numbers. do not put the minus sign outside. that would make the answer negative.

  46. anonymous
    • 5 years ago
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    welcome.

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