baby456
  • baby456
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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baby456
  • baby456
@jim_thompson5910
baby456
  • baby456
@zepdrix
zepdrix
  • zepdrix
Eyyy, let's not start with a number, let's start with .... "a number", say \(\large\rm x\).

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zepdrix
  • zepdrix
Double it: \(\large\rm 2x\)
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
Cube root the result: \(\large\rm \sqrt[3]{2x-3}\)
zepdrix
  • zepdrix
Add 4: \(\large\rm \sqrt[3]{2x-3}+4\) And we need to find the inverse of this? :o Is that the idea?
zepdrix
  • zepdrix
\[\large\rm y=\sqrt[3]{2x-3}+4\]What do you think baby? +_+ Were the steps so far a little confusing?
baby456
  • baby456
i get it now thanls.
baby456
  • baby456
i need more help because we have to do the opposite of the function.
zepdrix
  • zepdrix
So we have to "undo" every step that we made, and we have to do it in the reverse order.
zepdrix
  • zepdrix
The last thing we did was `add 4`. So to find our inverse our FIRST step will be to `subtract 4`
baby456
  • baby456
so would it be x-4
zepdrix
  • zepdrix
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?
zepdrix
  • zepdrix
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,\[\large\rm \color{orangered}{(x-4)^3}=2y-3\]
zepdrix
  • zepdrix
What next? Any ideas? :)
baby456
  • baby456
divide by 3
zepdrix
  • zepdrix
Hmm the 3 is being `subtracted`. Opposite of that would be to `add 3`, yes? :)
baby456
  • baby456
add 3
zepdrix
  • zepdrix
\[\large\rm \color{orangered}{(x-4)^3+3}=2y-3+3\]\[\large\rm \color{orangered}{(x-4)^3+3}=2y\]Oooo yay! We're almost there.
zepdrix
  • zepdrix
What next? We have a 2 `multiplying` our y. How do we undo that? :d
baby456
  • baby456
divide each side by 2
zepdrix
  • zepdrix
\[\large\rm \color{orangered}{\frac{(x-4)^3+3}{2}}=\frac{\cancel2y}{\cancel2}\]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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