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anonymous

  • one year ago

Let f(x)=3+×^2+tan((pi)(x)/2), where -1 <×<1 A) find f^-1 (3) B) find f (f^-1 (5))

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  1. anonymous
    • one year ago
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    I know that when you inverse a function you switch the x and the y values. But I don't know what to do with the y value instead of x in tangent.

  2. rational
    • one year ago
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    for part A, you need to solve : \[3 = 3+x^2+\tan(\pi x/2)\]

  3. anonymous
    • one year ago
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    Wouldn't the x value be 3?

  4. rational
    • one year ago
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    nope, as the name says, \(f^{-1}(3)\) is the value of \(x\) that makes the function \(f(x)\) spit out \(3\)

  5. rational
    • one year ago
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    \[f^{-1}(3) = a \implies f(a) = 3\] so you need to solve \(f(a)=3\) for \(a\)

  6. anonymous
    • one year ago
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    Ohhhh

  7. rational
    • one year ago
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    \[3 = 3+x^2+\tan(\pi x/2)\] \[0=x^2+\tan(\pi x/2)\] you can eyeball the value of \(x\) that satisfies above equation

  8. rational
    • one year ago
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    Yes!

  9. rational
    • one year ago
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    so \(f(0) = 3 \implies f^{-1}(3) = 0\)

  10. anonymous
    • one year ago
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    Wow so much simpler than I was doing it before! Thank you

  11. rational
    • one year ago
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    np what about part B

  12. rational
    • one year ago
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    hold up, we don't really need to find \(f^{-1}(5)\)

  13. anonymous
    • one year ago
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    And then solve for the inverse and then place into the original

  14. rational
    • one year ago
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    \(f\) eats \(f^{-1}\)

  15. rational
    • one year ago
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    \[\require{cancel} \large { f(f^{-1}(5))\\~\\ \cancel{f}(\cancel{f^{-1}}(5))\\~\\5 }\]

  16. anonymous
    • one year ago
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    Are you sure they cancel?

  17. ikram002p
    • one year ago
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    they dont cancel they generate new function f(f^-1(x))=x

  18. rational
    • one year ago
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    Think of it like this : \(f^{-1}(5)\) is the value of \(x\) that makes \(f(x)\) spit out \(5\), so if you feed that value to \(f(x)\) again, it simply spits out \(5\)

  19. rational
    • one year ago
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    |dw:1439073410822:dw|

  20. anonymous
    • one year ago
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    Ok

  21. rational
    • one year ago
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    |dw:1439073520821:dw|

  22. rational
    • one year ago
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    ofcourse assuming the inverse exists and one-to-one etc..

  23. rational
    • one year ago
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    Notice that in part B, you are inputting \(5\) to \(f^{-1}(x)\) (bottom box), it spits out \(x\), then you're inputting that value to \(f(x)\) again (top box) it spits out \(5\).

  24. anonymous
    • one year ago
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    Ok that makes sense. Thank you for your help

  25. rational
    • one year ago
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    np

  26. Loser66
    • one year ago
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    I would like to know how to solve \(x^2 + tan (\pi x/2)=0\) . @rational please.

  27. rational
    • one year ago
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    plugin x = 0 and see that it evaluates to 0 i don't think there is a way to solve it analytically

  28. Loser66
    • one year ago
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    ok, I got you. Since It is trivial when we graph them out. I would like to know whether we can solve it algebraically or not. Thanks for clarifying.

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