Barrelracing
  • Barrelracing
When 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 x minus four, 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. Who is correct? Justify your reasoning.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Barrelracing
  • Barrelracing
@jim_thompson5910
Barrelracing
  • Barrelracing
@freckles
jim_thompson5910
  • jim_thompson5910
` 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 x minus four` what a mess... hopefully I translated the word problem into the proper function \[\Large f(x) = \frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}\] let me know if I did or not

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Barrelracing
  • Barrelracing
yes you did
jim_thompson5910
  • jim_thompson5910
ok great
jim_thompson5910
  • jim_thompson5910
`Bella says that the function is defined at x = –1, x = 2, and x = 4` let's check this claim. If we were to replace EVERY x with a value like x = 4, then... \[\Large f(x) = \frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}\] \[\Large f(4) = \frac{(4-1)(4+2)(4+4)}{(4+1)(4-2)(4-4)}\] \[\Large f(4) = \frac{(3)(6)(8)}{(5)(2)(0)}\] \[\Large f(4) = \frac{144}{0}\] \[\Large f(4) = \text{Undefined}\] the result is undefined because we cannot divide by zero. We can't have 0 in the denominator
jim_thompson5910
  • jim_thompson5910
`Bella says that the function is defined at x = –1, x = 2, and x = 4` so Bella is wrong. At least about the `x=4` part. I recommend you check `x=-1` and `x=2`
Barrelracing
  • Barrelracing
ok thank you
jim_thompson5910
  • jim_thompson5910
no problem
Barrelracing
  • Barrelracing
the rest are not undefined
jim_thompson5910
  • jim_thompson5910
correct,x=-1 and x=2 are also undefined because they make the denominator 0
jim_thompson5910
  • jim_thompson5910
oh wait, you should say "the rest are undefined" or "the rest are not defined"
Barrelracing
  • Barrelracing
i got f(-1)=-4 and f(2)= 11
jim_thompson5910
  • jim_thompson5910
you did something wrong
Barrelracing
  • Barrelracing
f(-1) = (-1 - 1)(-1 + 2)(-1 + 4)/ (-1 + 1)(-1 - 2)(-1 - 4) = -2 + 1 + 3/ 0 - 3 - 5 = 2/-8 f(-1) = -4
Barrelracing
  • Barrelracing
f(2) = (2 - 1)(2 + 2)(2 + 4)/ (2 + 1)(2 - 2)(2 - 4) = 1 + 4 + 6/ 3 + 0 - 2 = 11/1 = 11
Barrelracing
  • Barrelracing
are you gonna help me still? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
sorry my notification thing isn't working properly
Barrelracing
  • Barrelracing
thats ok
jim_thompson5910
  • jim_thompson5910
(-1 + 1)(-1 - 2)(-1 - 4) = 0*(-3)*(-5) = 0 is in the denominator
jim_thompson5910
  • jim_thompson5910
the parenthesis next to each other means multiply
jim_thompson5910
  • jim_thompson5910
make sense?
Barrelracing
  • Barrelracing
ohhh ok thank you so much for the help.
jim_thompson5910
  • jim_thompson5910
no problem

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