anonymous
  • anonymous
f(x) = 2e^3x + 1 Find the inverse f^-1(x). Can anyone help?
Trigonometry
jamiebookeater
  • jamiebookeater
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jackellyn
  • jackellyn
Set it equal to y. Switch the x and the y, solve for y.
jackellyn
  • jackellyn
y=2x^3+1 x=2y^3+1 Solve for y.
anonymous
  • anonymous

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jackellyn
  • jackellyn
Did you get the answer?
anonymous
  • anonymous
my calculator is messing up, can you walk me through it pls. @jackellyn
jackellyn
  • jackellyn
so to solve for y you need to perform reverse operations. If we have x=2e^3y +1 What do we need to do first to get y alone?
anonymous
  • anonymous
subtract 3y on both sides @jackellyn i think
jackellyn
  • jackellyn
Well the 3y is part of the exponent so we can't get rid of it just yet. First we need to get rid of the +1 and move it to the other side so do we add it to both sides or subtract it?
anonymous
  • anonymous
subtract it @jackellyn
jackellyn
  • jackellyn
Right so we have x-1= 2e^3y Now we need to get rid of the 2 in front of the e^3y. What do you get?
anonymous
  • anonymous
divided
anonymous
  • anonymous
@jakellyn if thats correct
anonymous
  • anonymous
@jackellyn you there
jackellyn
  • jackellyn
Yes, divide so then we have (x-1)/2=e^3y Do you remember the opposite operation of e?
anonymous
  • anonymous
I don't recall @jackellyn
jackellyn
  • jackellyn
You need to take the natural log of both sides to get rid of the e and stay only with 3y. ln (x-1/2)=ln e^3y ln (x-1/2)=3y Solve for y.
anonymous
  • anonymous
\[y =\frac{ x }{ 3}-\frac{ 1 }{ 6 }\] is that correct? @jackellyn
anonymous
  • anonymous
so is this the inverse for \[f ^{-1}\]
anonymous
  • anonymous
@jackellyn idk if you saw my replies^^^^
jackellyn
  • jackellyn
Sorry, I didnt. You can't get rid of the ln. It stays all together and then you can divide the 3. So \[y=\frac{(\ln \frac{ x-1 }{ 2 }) }{ 3 }\]
anonymous
  • anonymous
so thats the inverse of \[f ^{-1}(x)\]
jackellyn
  • jackellyn
It is the inverse of f(x). \[f ^{-1}(x)\] is the notation for inverse.
anonymous
  • anonymous
okay i understand now thx a lot
anonymous
  • anonymous
@jackellyn one more question what is the inverse

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