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anonymous
 one year ago
When looking at a rational function, Charles and Bobby have two different thoughts. Charles says that the function is defined at x = 2, x = 3, and x = 5. Bobby says that the function is undefined at those x values. Describe a situation where Charles is correct, and describe a situation where Bobby is correct. Is it possible for a situation to exist that they are both correct? Justify your reasoning.
anonymous
 one year ago
When looking at a rational function, Charles and Bobby have two different thoughts. Charles says that the function is defined at x = 2, x = 3, and x = 5. Bobby says that the function is undefined at those x values. Describe a situation where Charles is correct, and describe a situation where Bobby is correct. Is it possible for a situation to exist that they are both correct? Justify your reasoning.

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Consider the function \(f(x) = \dfrac{1}{(x+2)(x3)(x5)}\) Is this function defined for \(x=2,3,\) or \(5\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes those are the vertical asymptotes? and the zeros/roots

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1NO! It is not defined for \(x=2, 3, \) and \(5\). What happens if you plug in 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when i graph the function on a graphing calculator the table says error, for 2,3,5. 3 doesnt work

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1right, \(\dfrac{1}{(2+3)(33)(53)}=\dfrac{1}{1*0*2}=\dfrac{1}{0}\) YIKES

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1So a function is not defined for a number that causes us to divide by \(0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so there is no time when the both of them can be correct??
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