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warpedkitten

  • one year ago

If f(x) = x^2 - 25 and g(x) = x - 5, what is the domain of (f/g)(x)?

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  1. UsukiDoll
    • one year ago
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    \[f(x) = x^2-25\] and \[g(x) = x-5 \] \[(\frac{f}{g})(x)\] is this your question and are those the right functions?

  2. warpedkitten
    • one year ago
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    Yes, that's correct.

  3. UsukiDoll
    • one year ago
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    alright so we need \[(\frac{f}{g})(x) \] \[(\frac{f}{g})(x) = \frac{x^2-25}{x-5}\] can you factor the \[x^2-25\]? after that step we can cancel a term out

  4. warpedkitten
    • one year ago
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    I don't know how to solve it, I'm not very good with math. lol.

  5. UsukiDoll
    • one year ago
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    oh ok. no problem . we can use the difference of squares formula \[(a^2-b^2) = (a+b)(a-b) \] so if we let a = x and b = 5 \[(x^2-5^2) = (x+5)(x-5) \]

  6. UsukiDoll
    • one year ago
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    so \[(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}\] so what does the numerator and denominator have in common

  7. UsukiDoll
    • one year ago
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    there's a x-5 in the numerator and denominator so that term is canceled out

  8. UsukiDoll
    • one year ago
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    \[(\frac{f}{g})(x) = x+5\] since our new function isn't a fraction , there are no restrictions so the domain is all real numbers.

  9. UsukiDoll
    • one year ago
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    the graph should be a straight line and it's a function because it passes the vertical line test. A vertical line test is needed to determine if a graph is a function or not. If it's a function then the vertical line should hit the graph only once. If it's not a function, the vertical line crosses the graph more than once.

  10. warpedkitten
    • one year ago
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    So my answer options are all real values of x all real values of x except x = 5 all real values of x except x = –5 all real values of x except x = 5 and x = –5 Would that mean it's a? Since they're all real numbers? Or would it be b?

  11. UsukiDoll
    • one year ago
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    has to be a... all reals. the function is not a fraction

  12. warpedkitten
    • one year ago
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    Thank you! :)

  13. UsukiDoll
    • one year ago
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    this is the graph of the function (f/g) (x)

  14. UsukiDoll
    • one year ago
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    if you draw a vertical line on this graph, the vertical line only touches once, so it's a function .

  15. zzr0ck3r
    • one year ago
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    hehe domain does not include x=5. It would not show up on a graph because its a hole. If you graph \(\frac{x^2-25}{(x-5)}\) and zoomed in infinitely close it would look like this |dw:1439266602373:dw| \[\frac{x^2-25}{(x-5)}\ne x+5\] in general (one is defined for 5, one is not) But if we remove \(5\) from the domain it is :)

  16. UsukiDoll
    • one year ago
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    if we factor the numerator, the x-5 cancels out leaving us with the new function no longer being a fraction. Hence, no restrictions... all reals. \[(\frac{f}{g})(x) = \frac{x^2-25}{x-5} \] \[(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5}\] \[(\frac{f}{g})(x) = x+5 \]

  17. UsukiDoll
    • one year ago
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    if you're not factoring then there is a restriction which is all reals except when x = 5. But I think that there have been times when canceling terms can happen.

  18. UsukiDoll
    • one year ago
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    both \[(\frac{f}{g})(x) = \frac{(x+5)(x-5)}{x-5} \] and \[(\frac{f}{g})(x) = x+5 \] produce different results.. It's like do we consider the factored version in that case it's all reals on the domain or the non-factored case all reals except x = 5 ?

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