Inverse of a Rational Function

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D!ddy

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Nov 2, 2007
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Hi, im new here but in my assignment i have this rational question that i have the find the inverse equation for.., i usually subsitute x for y before i start breaking it down but i saw somewhere in my textbook to do it last, so i did it last and checked on my calculator (ti-83 plus) through drawinvY1 trick but on my Y2(equation i found at the end) it just gets bolded..i dunno if thats correct because if i had it right i wouldnt see any changes...anyways heres my equation
f(x)=(2X+1)\(x+2)
i found
y=(1-2x)\(x-2)
 
To check if a function is indeed the inverse of another, plug one into the other, and see if you end up with just x. In this case, try plugging f<sup>-1</sup> into f:

. . .\(\displaystyle \L f(f^{-1}(x))\)

. . . . .\(\displaystyle \L =\,\frac{2\left({\frac{1\, -\, 2x}{x\, -\, 2}}\right)\, +\, 1}{\left(\frac{1\, -\, 2x}{x\, -\, 2}\right)\, +\, 2}\)

. . . . .\(\displaystyle \L =\,\frac{\left(\frac{2(1\,-\,2x)}{x\,-\,2}\,+ \,\frac{x\,-\,2}{x\,-\,2}\right)}{\left(\frac{1\,-\,2x}{x\,-\,2}\,+\, \frac{2(x\,-\,2)}{x\,-\,2}\right)}\)

. . . . .\(\displaystyle \L =\,\frac{\left(\frac{2\,-\,4x\,+\,x\,-\,2}{x\,- \,2}\right)}{\left(\frac{1\,-\,2x\,+\,2x\,-\,4}{x\,-\,2}\right)}\)

. . . . .\(\displaystyle \L =\,\frac{\left(\frac{-3x}{x\,-\,2}\right)} {\left(\frac{-3}{x\,-\,2}\right)}\)

. . . . .\(\displaystyle \L =\,\frac{-3x}{-3}\)

. . . . .\(\displaystyle \L =\,x\)

This confirms your inverse. :D

Eliz.
 
WoW not only did you answer my question you taught me a cool way to check my answer thanks i wont forget that trick
 
on that same question they ask, "On which line do points that are common to the graphs of both functions lie?" is it asking for the invariant points? and if it is, can you tell me what words stood out to tell you that, because its all confusing why dont they just ask you " What are the invariant points"??? thanks
 
D!ddy said:
WoW not only did you answer my question you taught me a cool way to check my answer thanks i wont forget that trick
Click to expand...

It's not a trick
 
You'll need to check your book for what they mean by "invariant points", but I suspect they're wanting to know what points lie on both graphs. These points, naturally, would be the intersection points of the graphs, which can be found by setting the functions equal to each other. (A quick graph on your calculator should serve to confirm or correct whatever answer you get.)

Eliz.
 
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