anonymous
  • anonymous
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.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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zzr0ck3r
  • zzr0ck3r
Consider the function \(f(x) = \dfrac{1}{(x+2)(x-3)(x-5)}\) Is this function defined for \(x=-2,3,\) or \(5\)?
zzr0ck3r
  • zzr0ck3r
and why?
anonymous
  • anonymous
yes those are the vertical asymptotes? and the zeros/roots

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zzr0ck3r
  • zzr0ck3r
NO! It is not defined for \(x=-2, 3, \) and \(5\). What happens if you plug in 3?
anonymous
  • anonymous
when i graph the function on a graphing calculator the table says error, for -2,3,5. 3 doesnt work
zzr0ck3r
  • zzr0ck3r
right, \(\dfrac{1}{(-2+3)(3-3)(5-3)}=\dfrac{1}{1*0*2}=\dfrac{1}{0}\) YIKES
zzr0ck3r
  • zzr0ck3r
So a function is not defined for a number that causes us to divide by \(0\).
anonymous
  • anonymous
so there is no time when the both of them can be correct??

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