anonymous
  • anonymous
how to you find the range of f(x)=x-4/x^2-4 PLEASE HELP
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{x-4}{x^2-4}\]?
anonymous
  • anonymous
calculus problem or something else?
anonymous
  • anonymous
guess i'll never know...

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anonymous
  • anonymous
sorry! its pre cal and yest thats correct!
anonymous
  • anonymous
ok without calc it is not going to be that easy
anonymous
  • anonymous
do you have any insight on how to solve i t?
anonymous
  • anonymous
yes, and it is a pain
anonymous
  • anonymous
we are going to have to solve for x
anonymous
  • anonymous
\[y=\frac{x-4}{x^2-4}\] \[yx^2-4y=x-4\] \[yx^2-x-4y-4=0\] now we have a quadratic equation in x, so we can solve using the quadratic formula
Mertsj
  • Mertsj
\[x=\frac{1\pm \sqrt{1+16y^2+16y}}{2y}\]
anonymous
  • anonymous
@Mertsj if you have a snap way to do it let me know
anonymous
  • anonymous
ok we are doing the same thing
Mertsj
  • Mertsj
Graph it.
anonymous
  • anonymous
denominator is y^2 but it makes no difference now you have to solve \[16x^2+16x+1\geq 0\] lord!
anonymous
  • anonymous
oh no you are right, denominator is 2y
Mertsj
  • Mertsj
How so? a = y doesn't it?
anonymous
  • anonymous
so we have to solve [16x^2+16y+1=0\] using the quadratic formula, another pain oh yes you are right, i am wrong about the denominator
anonymous
  • anonymous
\[16x^2+16y+1=0\]
anonymous
  • anonymous
\[x=\frac{-2\pm\sqrt{3}}{4}\]
Mertsj
  • Mertsj
I got 8 on the bottom
anonymous
  • anonymous
also wrong, sorry range is \[(-\infty,\frac{-2-\sqrt{3}}{4}]\cup (\frac{-2+\sqrt{3}}{4},\infty)\]
anonymous
  • anonymous
well i have to admit i cheated for the last part http://www.wolframalpha.com/input/?i=16x^2%2B16x%2B1%3E0
Mertsj
  • Mertsj
I know. I found it.
Mertsj
  • Mertsj
Do you really think a pre-calc student was supposed to do that?

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