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anonymous

  • one year ago

Looking at the rational function f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity x plus one times the quantity x minus two times the quantity Bella says that the function is defined at x = –1, x = 2, and x = 4. Edward says that the function is undefined at those x values. Who is correct? Justify your reasoning.

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  1. phi
    • one year ago
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    writing equations out in English makes it very hard to read.

  2. phi
    • one year ago
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    can you use the equation editor?

  3. anonymous
    • one year ago
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    how do I do that?

  4. phi
    • one year ago
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    click on the equation button in the lower left of the input area

  5. anonymous
    • one year ago
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    \[f(x)=(x-1)(x+2)(x+4)/ (x+1)(x-2)(x-4)\]

  6. anonymous
    • one year ago
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    I couldn't find the "divide" button hahah

  7. phi
    • one year ago
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    the fraction button is there, last row, on the right

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  8. phi
    • one year ago
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    anyway, you hopefully know we do not allow "divide by 0" It is not done. in your problem, if you make x certain values, the "bottom" will have a zero and zero times anything is zero and you will have stuff/0 and we don't allow that.

  9. anonymous
    • one year ago
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    @peachpi

  10. anonymous
    • one year ago
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    Who would be right??

  11. phi
    • one year ago
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    Do you understand that if you have (for example) (x+1) in the bottom what number (for x) makes x+1 be 0 ?

  12. anonymous
    • one year ago
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    -1

  13. phi
    • one year ago
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    and if x=-1 we will get in the bottom 0 * (-1-2)*(-1-4) or 0 * -3 * -5 notice the -3 and -5 don't matter, because when we multiply it out we get 0 *0 in the bottom* is not allowed

  14. phi
    • one year ago
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    to prevent the divide by 0, we do not allow x=-1 then what is the answer for that expression when x=-1? answer: we don't have an answer. It is undefined. notice there are two other values that make the bottom 0 we don't allow those either. can you take a guess as to who is right for this question?

  15. anonymous
    • one year ago
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    edward!

  16. anonymous
    • one year ago
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    Yes?

  17. phi
    • one year ago
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    yes.

  18. anonymous
    • one year ago
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    Thanks<3

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