Question

Help me please!!!

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 where they are both correct? Justify your reasoning.

Answer 1

When you plug in -2,3, or 5, you get 0/0, which is mathematical garbage, so we don't define the function at that point. However, carelessly dividing out the numerator and denominator without remembering the special cases of x=-2,3,5 will result in f(x)=1 everywhere including x=-2,3,5 which is not quite correct

Answer 2

\[f(x)=\frac{(x+2)(x-3)(x-5)}{(x+2)(x-3)(x-5)}\]

Answer 3

I am not understanding how to do it.

Answer 4

when you plug in x= 16 for\[\frac{x-16}{x-16}\]what do you get?

Answer 5

0

Answer 6

You get\[\frac{0}{0}\]Which makes no sense at all. So we can't say that the function at x=16 is 1

Answer 7

\[\frac{0}{0}\] is not equal to any number

Answer 8

Yes I understand that

Answer 9

Ok, so expanding on the idea that where a function has a place where there can be a \[\frac{0}{0}\]we can combine multiple of these things that produce \[\frac{0}{0}\]

Answer 10

The answer will be.....

When you plug in those values, it equals 0/0 which is undefined, which makes Bobby correct. To make Charles correct, you have to_____________.

I dont know how Charles is.

Answer 11

Charles could be correct by remembering that x=-2,3,5 the function evaluates to \[\frac{0}{0}\] and so knowing it's not right to say the function exists at those points, he should agree with Bobby that the function does not exist at those points

Answer 12

Thank you!!!