anonymous
  • anonymous
Domain & Range HELP! Please help with two functions. Click here and see the coming soon drawing...UPDATE: Drawing is now available!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1347943151983:dw|
anonymous
  • anonymous
domain of sine is all real numbers
anonymous
  • anonymous
range of sine is \([-1,1]\) so range of -3 sine is \([-3,3]\) no algebra needed for that one

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anonymous
  • anonymous
oh i didn't see the \(+1\) out at the end. adjust by adding one to get range of \([-2,4]\)
anonymous
  • anonymous
How do I find the domain of the sine equation?
anonymous
  • anonymous
domain of sine is all real numbers, so unless there is some restriction inside, there is no restriction
anonymous
  • anonymous
domain of \(2x^4-5\) is all reals, so no worries here
anonymous
  • anonymous
Okay... What about the second one? I know that the denominator cant equal 0, but what about range??
anonymous
  • anonymous
for range find the range of \[\frac{x-2}{x+1}\] and then exclude what you would get if \(x=1\)
anonymous
  • anonymous
since in the original function you know \(x\neq 1\)
anonymous
  • anonymous
I'm having difficulty finding the range..
anonymous
  • anonymous
this thing can never be 1 because a fraction is only one if the numerator and denominator are equal, and in this case they are not
anonymous
  • anonymous
other than that, it is all real numbers. so you have only to exclude two values from the range. it cannot be 1 and it cannot be \[\frac{1-2}{1+1}=-\frac{1}{2}\] because you are not allowed to evaluate at \(x=1\)
anonymous
  • anonymous
if you want some more math to do to make your teacher happy, solve for \(x\) in \[y=\frac{x-2}{x+1}\]
anonymous
  • anonymous
or switch \(x\) and \(y\) and solve for \(y\), either way will do it can you do that algebra?
anonymous
  • anonymous
I think I got it now, @satellite73! Thanks so much! I have a Functions Test tomorrow!
anonymous
  • anonymous
good luck!

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