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Study23

  • 3 years ago

Domain & Range HELP! Please help with two functions. Click here and see the coming soon drawing...UPDATE: Drawing is now available!!!

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  1. Study23
    • 3 years ago
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    |dw:1347943151983:dw|

  2. anonymous
    • 3 years ago
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    domain of sine is all real numbers

  3. anonymous
    • 3 years ago
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    range of sine is \([-1,1]\) so range of -3 sine is \([-3,3]\) no algebra needed for that one

  4. anonymous
    • 3 years ago
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    oh i didn't see the \(+1\) out at the end. adjust by adding one to get range of \([-2,4]\)

  5. Study23
    • 3 years ago
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    How do I find the domain of the sine equation?

  6. anonymous
    • 3 years ago
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    domain of sine is all real numbers, so unless there is some restriction inside, there is no restriction

  7. anonymous
    • 3 years ago
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    domain of \(2x^4-5\) is all reals, so no worries here

  8. Study23
    • 3 years ago
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    Okay... What about the second one? I know that the denominator cant equal 0, but what about range??

  9. anonymous
    • 3 years ago
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    for range find the range of \[\frac{x-2}{x+1}\] and then exclude what you would get if \(x=1\)

  10. anonymous
    • 3 years ago
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    since in the original function you know \(x\neq 1\)

  11. Study23
    • 3 years ago
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    I'm having difficulty finding the range..

  12. anonymous
    • 3 years ago
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    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

  13. anonymous
    • 3 years ago
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    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 \(x=1\)

  14. anonymous
    • 3 years ago
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    if you want some more math to do to make your teacher happy, solve for \(x\) in \[y=\frac{x-2}{x+1}\]

  15. anonymous
    • 3 years ago
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    or switch \(x\) and \(y\) and solve for \(y\), either way will do it can you do that algebra?

  16. Study23
    • 3 years ago
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    I think I got it now, @satellite73! Thanks so much! I have a Functions Test tomorrow!

  17. anonymous
    • 3 years ago
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    good luck!

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