anonymous
  • anonymous
I'll give a medal and fan who answers this When looking at a rational function, Bella and Edward have two different thoughts. 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. Describe a situation where Bella is correct, and describe a situation where Edward is correct. Is it possible for a situation to exist where they are both correct? Justify your reasoning
Mathematics
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anonymous
  • anonymous
I'll give a medal and fan who answers this When looking at a rational function, Bella and Edward have two different thoughts. 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. Describe a situation where Bella is correct, and describe a situation where Edward is correct. Is it possible for a situation to exist where they are both correct? Justify your reasoning
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
ok, I give you the simplest form to Bella, she said the functions is defined at x = -1, x =2 and x =4, her function should be \[f(x) = \frac{x^2-6x+8}{(x^2+1)(x-2)(x-4)} \] because when we factor the numerator, it cancel the last 2 terms of denominator and it turns to \[f(x)=\dfrac{1}{x^2+1} \] which is defined in all real number.
anonymous
  • anonymous
to Edward, just let the numerator =1, and (x-1) instead of (x^2-1) in denominator, the function turns to undefined at those points
anonymous
  • anonymous
They cannot be correct at the same time. That's what I thought

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