anonymous
  • anonymous
PLEASe HELP GIVE MEDAL! Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = x/-7/x+2 and g(x) = -2x-7/x-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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DebbieG
  • DebbieG
\(\Large f(x)=\dfrac{ x-7 }{ x+2 }\) and \(\Large g(x)=\dfrac{-2x-7 }{ x-1 }\) So to do the composition f(g(x)) you just drop that (g(x)) into f(x), where you see x: \(\Large f(g(x))=\dfrac{ \dfrac{-2x-7 }{ x-1 }-7 }{ \dfrac{-2x-7 }{ x-1 }+2 }\) To simplify this, I would start by multiplying num'r and den'r by (x-1): \[\left( d \right)\] \(\Large \left( \dfrac{ \dfrac{-2x-7 }{ x-1 }-7 }{ \dfrac{-2x-7 }{ x-1 }+2 }\right)\cdot\dfrac{x-1}{x-1}\) I know this looks hairy, but just take it one step at a time..... notice that the "inner" den'rs cancel nicely, then you just distribute the (x-1) to the other term: \(\Large \dfrac{ -2x-7 -7(x-1) }{ -2x-7 +2(x-1) }=\dfrac{ -2x-7 -7x+7 }{ -2x-7 +2x-2 }\) \(\Large =\dfrac{ -9x }{ -9 }=x\) Kinda like magic.... you get back x!! :) Now that's only the first direction, f(g(x))=x... now you can try the other direction, to show that g(f(x))=x also. Then that proves that they are inverses.
anonymous
  • anonymous
ok ill try
anonymous
  • anonymous
ok i dnt get it

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DebbieG
  • DebbieG
This time you are droppin f(x) into g(x) for the x's.... so you start here: \(\Large g(f(x))=\dfrac{-2\cdot\dfrac{ x-7 }{ x+2 }-7 }{\dfrac{ x-7 }{ x+2 }-1 }\) Did you do that? What's next?
anonymous
  • anonymous
ok hold on
anonymous
  • anonymous
im writing it on a piece of paper
DebbieG
  • DebbieG
I would hope so, lol. :) HINT: look up top, the first thing I did was to clear those "inner fractions" from the big fraction, by multiplying num'r and den'r by the den'r of (x-1). This time, you'll want to multiply both by (x+2).
anonymous
  • anonymous
okay i dont think i can do this
anonymous
  • anonymous
what do i do with the -2 in the front
DebbieG
  • DebbieG
Did you clear the fractions by multiplying by (x+2)/(x+2)? As for the -2, remember that when there is a number multiplied by a fraction, you can think of the num'r of that fraction as having "invisible" ( )'s around it, so you'll just distribute through the num'r: \(\Large \left(\dfrac{-2\cdot\dfrac{ x-7 }{ x+2 }-7 }{\dfrac{ x-7 }{ x+2 }-1 }\right)\cdot \dfrac{x+2}{x+2}=\dfrac{-2( x-7 )-7(x+2) }{ x-7 -1(x+2) }\) Does that make sense? Now see if you can finish it up from there. Just distribute the -2 through the first ( ) in the num'r; distribute the -7 through the 2nd one; and distribute the -1 through the ( ) in the den'r. Simplify num'r and den'r by combining like terms, and it should all fall into place!

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