anonymous
  • anonymous
find f^-1(x) for a) y= 3x-2/x+4 b) f(x)= (sqaureroot) x-1
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
How is the first equation set up?\[y=(3x-2)/(x+4)\]or\[y=3x-(2/x)+4\]
anonymous
  • anonymous
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anonymous
  • anonymous
\[f ^{-1}(x)=f(y)\]\[x=(3y-2)/(y+4)\]Solve for y.\[(y+4)(x)=((3y-2)/(y+4))(y+4)\]\[(y+4)(x)=3y-2\]-3y -3y\[(y+4)(x)-3y=-2\] /x /x\[y+4-3y=-2/x\]\[-2y+4=-2/x\] -4 -4\[-2y=(-2/x)-4\]/-2 /-2\[y=((-2/x)-4)/-2\]It'll look ugly only because it's an inverse function. Although, after graphing it I may be wrong altogether.

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anonymous
  • anonymous
\[f^{-1}(x)=f(y)\] \[x=(3y−2)/(y+4)\]\[x(y+4)=(3y−2)\]\[xy+4x-3y+2=0\]\[y(x-3)+(4x+2)=0\]\[y(x-3)=-(4x+2)\]\[y=\frac{4x+2}{3-x}\]
anonymous
  • anonymous
For b, \[x=\sqrt{y-1}\]\[x^2=y-1\]\[y=x^2+1\]
anonymous
  • anonymous
thank you so much both of you (:
anonymous
  • anonymous
but im confused both of you got a different answer for A
anonymous
  • anonymous
He's right, I didn't distribute my x in one of my steps.
anonymous
  • anonymous
ohhh ok thanks

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