danmac0710
  • danmac0710
Is there any algebraic method for finding the range of a rational function such as: f(x)=(x-7)/(x-2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Hero
  • Hero
Find the inverse of the function.
amistre64
  • amistre64
y = (x-7)/(x-2) (x-2-5)/(x-2) 1 - 5/(x-2) x < 2 and 2 < x x-2 < 0 and 0 < x-2 1/(x-2) < 0 and 0 < 1/(x-2) -5/(x-2) < 0 and 0 < -5/(x-2) 1-5/(x-2) < 1 and 1 < 1-5/(x-2) y < 1 and 1 < y
amistre64
  • amistre64
might have an error abounding :)

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amistre64
  • amistre64
the horizontal asymptote is 1, so im satisfied
Hero
  • Hero
\[y = \frac{x - 7}{x - 2}\] Swap x and y \[x = \frac{y - 7}{y - 2}\] Then isolate y: \[x(y - 2) = y - 7\] \[xy - 2x = y - 7\] \[xy - y = 2x - 7\] \[y(x - 1) = 2x - 7\] \[y = \frac{2x - 7}{x - 1}\] Thus \[y \ne 1\]
Hero
  • Hero
@amistre64 what are you hinting at?
Hero
  • Hero
With your "error aboundings". Cool it with that stuff
amistre64
  • amistre64
i was sure i made a mistake in typing :)
Hero
  • Hero
To find the range, find the inverse of the function. The domain of the inverse is the range of the original function.
amistre64
  • amistre64
and your write up is fine, but there was no need to swap x for y; in the end your result was x\(\ne\) 1
Hero
  • Hero
I just addressed that just a second ago
amistre64
  • amistre64
there is more than 1 algebraic way to find the range. I simply presented one of them.
Hero
  • Hero
Is there a specific name for that method? I know you can find the inverse of the function or you can take the limit as x goes to infinity.
amistre64
  • amistre64
lol, im not good with names
Hero
  • Hero
I can't keep track of methods if they don't have a name.
Hero
  • Hero
@amistre64, just out of curiousity, what method would you use to find the range of this: \[f(x) = \frac{\sqrt{2x - 5}}{x - 3}\]
amistre64
  • amistre64
i cant find a name for what i did :) and asymptote is a thought; bottom is a higher degree than the top, so it goes to 0.
amistre64
  • amistre64
yeah, theres no easy way to do an inequality run with that one
Hero
  • Hero
There's a way to find the inverse, but it is quite tedious.
DebbieG
  • DebbieG
For \(\large f(x) = \dfrac{\sqrt{2x - 5}}{x - 3}\), can't we just use the facts that: 1. The domain is \(x \ge\dfrac{5}{2}\) 2. there is a vert asymptote at x=3 3. \(y\rightarrow\infty\) as \(x\rightarrow3^{+}\) and \(y\rightarrow -\infty\) as \(x\rightarrow 3^{-}\) 4. \(f(\dfrac{5}{2})=0\) 5. as \(x\rightarrow\infty\), \(y\rightarrow0\), because the degree of the num'r> degree of den'r. 6. The quotient of 2 continuous functions is continuous everywhere that the den'r is nonzero. So putting that alllll together, we get range = \((-\infty,\infty)\), right? I'm sure I'm missing something, lol.
amistre64
  • amistre64
domain is\[\{\frac52\le x<3\}\cup \{3
DebbieG
  • DebbieG
oh, of course it is.... sorry, I did know that but was sloppy above. lol
DebbieG
  • DebbieG
Should have said \(x \ge\dfrac{5}{2},~~x \neq3\)
amistre64
  • amistre64
hmm, y=0 for x=5/2 and then it would run up to x=3 in the y=-inf then run from x=3 from y=+inf and then horizontal out at y=0 which was already accounted for at the start of it so i like the -inf to inf rande :) just not very algebraic
DebbieG
  • DebbieG
Yeah, that's basically what I was getting at... we can show that on [5/2,3) the range is \((-\infty,0]\) and on \((3,\infty)\) the range is \((0,\infty)\), so the union gives the whole range then as \((-\infty,\infty)\). But I'm not sure if it is rigorous enough for what Hero was looking for...? :)
amistre64
  • amistre64
... and it has to have a name :)
DebbieG
  • DebbieG
LOL right, that too! :)

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