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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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.

Mathematics
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Eyyy, let's not start with a number, let's start with .... "a number", say \(\large\rm x\).

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Other answers:

Double it: \(\large\rm 2x\)
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.
Cube root the result: \(\large\rm \sqrt[3]{2x-3}\)
Add 4: \(\large\rm \sqrt[3]{2x-3}+4\) And we need to find the inverse of this? :o Is that the idea?
\[\large\rm y=\sqrt[3]{2x-3}+4\]What do you think baby? +_+ Were the steps so far a little confusing?
i get it now thanls.
i need more help because we have to do the opposite of the function.
So we have to "undo" every step that we made, and we have to do it in the reverse order.
The last thing we did was `add 4`. So to find our inverse our FIRST step will be to `subtract 4`
so would it be x-4
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?
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,\[\large\rm \color{orangered}{(x-4)^3}=2y-3\]
What next? Any ideas? :)
divide by 3
Hmm the 3 is being `subtracted`. Opposite of that would be to `add 3`, yes? :)
add 3
\[\large\rm \color{orangered}{(x-4)^3+3}=2y-3+3\]\[\large\rm \color{orangered}{(x-4)^3+3}=2y\]Oooo yay! We're almost there.
What next? We have a 2 `multiplying` our y. How do we undo that? :d
divide each side by 2
\[\large\rm \color{orangered}{\frac{(x-4)^3+3}{2}}=\frac{\cancel2y}{\cancel2}\]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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