Question

MEDAL REWARDED PLEASE HELP

Answer 1

@ganeshie8

Answer 2

let's take a peek
\(\huge g(f(x)) = \cfrac{1}{x-2} \)

now, let's give "x" some values, that is, DOMAIN
let's give x = 2
so what would our fraction become?

Answer 3

1/1

Answer 4

1/1
let's see
\(\huge g(f(x)) = \cfrac{1}{2-2} \ne \cfrac{1}{1}\)

Answer 5

so what do i do

Answer 6

i think it b or c

Answer 7

well, we'd like to find what will the fraction become if we use x = 2

Answer 8

what

Answer 9

say let's give x =2 in \(\huge \cfrac{1}{x-2}\)
what do we get?

Answer 10

1

Answer 11

1, let's see \(\huge \cfrac{1}{2-2} = \cfrac{1}{0} \ne 1\)

Answer 12

we just did this..

Answer 13

@Jhannybean @e.mccormick @tcarroll010

Answer 14

Did you get this one finally?

Answer 15

For the domain question, another way that sometimes put it is, "What are the domain restrictions?" If you can find the restrictions, then the domain is everything OTHER than those restrictions.

There are two general restrictions for ANY question that involves real numbers:
\(\Large \cfrac{a}{0}\) division by zero is not allowed.
\(\Large \sqrt[n]{a}\) if the nth root is an even root, like the square root or the 4th root and so on, then a must be non-negative, AKA: \(a\ge 0\)

In this particular case, there are no roots so we don't have to worry about that one. I just point it out for any other problems you do.

Therefore you are left with just a couple steps. First, find the equation that \(G(F(x))\) makes. To do this, you are putting the \(F(x)\) inside of \(G(y)\) so it replaces the y. That will get you some equation with a fraction. Now, find what values would set the bottom of the fraction to 0. Because 0 is NOT allowed, this tells you the restricted value(s). The domain is everything other than those restricted value(s).

Answer 16

Wow long explanation. For these types of questions you find the number(s) that make the function discontinuous.

A discontinuous function is found when evaluating the denominator of the function for x that makes the function nonexistent at that point but continuous everywhere except at that point.

In this case, we are finding a vertical asymptote that would indicate our point if discontinuity.

Answer 17

Yes... long... and I tried to break the concepts down. Some people get long and explanatory. Some get concise and term laden. Never know which, so I tend to run long.