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anonymous

  • one year ago

graph and find the inverse of f(x)=2x^2-4. Once you find the inverse, graph it too.

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  1. zzr0ck3r
    • one year ago
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    This only has a restricted inverse since the function is not 1-1

  2. anonymous
    • one year ago
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    too bad it doesn't have an inverse ...

  3. zzr0ck3r
    • one year ago
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    and what I mean is that it does not have an inverse :)

  4. anonymous
    • one year ago
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    Hmm..so would I just graph f(x)=2x^2-4 and then say it doesnt have an inverse?

  5. zzr0ck3r
    • one year ago
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    You can say that by the definition of an inverse, it must be 1-1 and it is not.

  6. anonymous
    • one year ago
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    Thanks :)

  7. zzr0ck3r
    • one year ago
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    or I guess graph it and draw a horizontal line through any two points to show it is not 1-1

  8. anonymous
    • one year ago
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    What about y=-3x+6? I got -x+3/6 as the inverse

  9. zzr0ck3r
    • one year ago
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    hmm

  10. zzr0ck3r
    • one year ago
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    \(y=-3x+6\) Switch the \(x\) and the (y\) and solve for \(y\). \(x=-3y+6\\ x-6=-3y\\ \dfrac{x-6}{-3}=y\\ y=\dfrac{6-x}{3}\)

  11. anonymous
    • one year ago
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    not to butt in but you can still solve \[2y^2-4=x\]for \(y\) to find an inverse, it just won't be a function

  12. zzr0ck3r
    • one year ago
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    or restrict the domain on the first one

  13. anonymous
    • one year ago
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    add 4, divide by 2 and you get \[y^2=\frac{x+4}{2}\] but when you solve for \(y\) you get \[y=\pm\sqrt{\frac{x+4}{2}}\]

  14. anonymous
    • one year ago
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    the \(\pm\) make it not a function

  15. zzr0ck3r
    • one year ago
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    \(f:\mathbb{R}^+\rightarrow R, f(x) = 2x^2-4\) has as its inverse a proper function.

  16. zzr0ck3r
    • one year ago
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    I think they meant, use the graph to determine if it has an inverse.

  17. anonymous
    • one year ago
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    Thanks guys:D

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