f composed with g

Question

let f(x) = 1/x-3  and g(x)= 1/x-6

then the domain of f composed with g is equal to all the reals except for two values, a and b with a<b and 

a=

b= 

Answer 1

Hi Again Dalia;

It is good to hear from you.  You ask great questions!

In the equation of...

1/x-3

I am assuming this is...

1/(x-3)

If x=3, then this results in 1/0.  This is not a real number.

 

In the equation of...

1/x-6

I am assuming this is...

1/(x-6)

if x=6, then this results in 1/0.  This is not a real number.

a=3

b=6

 

I believe "f composed with g" is

f((g)x)

If this is only f(x), then x would be considered as x.

However this is f((g)x), then x must be substituted with g(x).

 

I believe this equation is now...

f((g(x))=[1/(1/(x-6))]-3

 

[1/(1/(x-6))]

is

        1          

        1   

    (x-6)

 

This must be flipped to yield (x-6)

f((g)x)=x-6-3=x-9

 

The domain is all real numbers except a & b, a<b, 3<6.

 

I am only 90% certain I am correct.  I hope I receive feedback from other Wyzants.  I know I am not always correct.  Their comments are very helpful, constructive criticisms.  I need all of it.

 

Please let me know...