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baby456

  • one year ago

please helppppppppppp! Inverse Function! me with this its about inverse functions my teacher gave me she wants us to write an inverse function its direction are to start with a number and double it subtract three from the number cube root the result and the add 4 please help i started with the number 20 and need help from this helppppp.

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  1. baby456
    • one year ago
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    @jim_thompson5910

  2. baby456
    • one year ago
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    @zepdrix

  3. zepdrix
    • one year ago
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    Eyyy, let's not start with a number, let's start with .... "a number", say \(\large\rm x\).

  4. zepdrix
    • one year ago
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    Double it: \(\large\rm 2x\)

  5. zepdrix
    • one year ago
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    Subtract 3 from "the number": \(\large\rm 2(x-3)\) or this \(\large\rm 2x-3\) depending on what teacher meant by "the number". That's very ambiguous :( Makes it hard to interpret.

  6. zepdrix
    • one year ago
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    Cube root the result: \(\large\rm \sqrt[3]{2x-3}\)

  7. zepdrix
    • one year ago
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    Add 4: \(\large\rm \sqrt[3]{2x-3}+4\) And we need to find the inverse of this? :o Is that the idea?

  8. zepdrix
    • one year ago
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    \[\large\rm y=\sqrt[3]{2x-3}+4\]What do you think baby? +_+ Were the steps so far a little confusing?

  9. baby456
    • one year ago
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    i get it now thanls.

  10. baby456
    • one year ago
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    i need more help because we have to do the opposite of the function.

  11. zepdrix
    • one year ago
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    So we have to "undo" every step that we made, and we have to do it in the reverse order.

  12. zepdrix
    • one year ago
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    The last thing we did was `add 4`. So to find our inverse our FIRST step will be to `subtract 4`

  13. baby456
    • one year ago
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    so would it be x-4

  14. zepdrix
    • one year ago
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    No let's start with the "function" that we ended up with after applying all of those steps. Or, yes. I suppose the first in finding an inverse function is to switch your y and x. So we start with this:\[\large\rm \color{orangered}{x}=\sqrt[3]{2y-3}+4\]And after we apply a bunch of steps, our right side will turn into y, and whatever we end up with on the left side, the orange, will be our inverse function. So like I was saying, we'll subtract 4 as our first step,\[\large\rm \color{orangered}{x-4}=\sqrt[3]{2y-3}+4-4\]\[\large\rm \color{orangered}{x-4}=\sqrt[3]{2y-3}\]Do you see how that got rid of the +4 on the right side?

  15. zepdrix
    • one year ago
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    The second to last step when we were building our function was to take `cube root`. The inverse of a root is an power. Since cube corresponds to 3rd root, we'll apply a 3rd power to each side.\[\large\rm \color{orangered}{(x-4)^3}=(\sqrt[3]{2y-3})^3\]The cube root and cube power undo one another,\[\large\rm \color{orangered}{(x-4)^3}=2y-3\]

  16. zepdrix
    • one year ago
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    What next? Any ideas? :)

  17. baby456
    • one year ago
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    divide by 3

  18. zepdrix
    • one year ago
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    Hmm the 3 is being `subtracted`. Opposite of that would be to `add 3`, yes? :)

  19. baby456
    • one year ago
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    add 3

  20. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{(x-4)^3+3}=2y-3+3\]\[\large\rm \color{orangered}{(x-4)^3+3}=2y\]Oooo yay! We're almost there.

  21. zepdrix
    • one year ago
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    What next? We have a 2 `multiplying` our y. How do we undo that? :d

  22. baby456
    • one year ago
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    divide each side by 2

  23. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{\frac{(x-4)^3+3}{2}}=\frac{\cancel2y}{\cancel2}\]Ok undo multiply by dividing, good good good.\[\large\rm \color{orangered}{\frac{(x-4)^3+3}{2}}=y\]Yayyyy good job \c:/ We did it!

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