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Barrelracing
 one year ago
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.
Barrelracing
 one year ago
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.

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Barrelracing
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1` 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 \[\Large f(x) = \frac{(x1)(x+2)(x+4)}{(x+1)(x2)(x4)}\] let me know if I did or not

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1`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... \[\Large f(x) = \frac{(x1)(x+2)(x+4)}{(x+1)(x2)(x4)}\] \[\Large f(4) = \frac{(41)(4+2)(4+4)}{(4+1)(42)(44)}\] \[\Large f(4) = \frac{(3)(6)(8)}{(5)(2)(0)}\] \[\Large f(4) = \frac{144}{0}\] \[\Large f(4) = \text{Undefined}\] the result is undefined because we cannot divide by zero. We can't have 0 in the denominator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1`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
 one year ago
Best ResponseYou've already chosen the best response.0the rest are not undefined

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1correct,x=1 and x=2 are also undefined because they make the denominator 0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1oh wait, you should say "the rest are undefined" or "the rest are not defined"

Barrelracing
 one year ago
Best ResponseYou've already chosen the best response.0i got f(1)=4 and f(2)= 11

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you did something wrong

Barrelracing
 one year ago
Best ResponseYou've already chosen the best response.0f(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
 one year ago
Best ResponseYou've already chosen the best response.0f(2) = (2  1)(2 + 2)(2 + 4)/ (2 + 1)(2  2)(2  4) = 1 + 4 + 6/ 3 + 0  2 = 11/1 = 11

Barrelracing
 one year ago
Best ResponseYou've already chosen the best response.0are you gonna help me still? @jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1sorry my notification thing isn't working properly

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1(1 + 1)(1  2)(1  4) = 0*(3)*(5) = 0 is in the denominator

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the parenthesis next to each other means multiply

Barrelracing
 one year ago
Best ResponseYou've already chosen the best response.0ohhh ok thank you so much for the help.
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