Find f(g(x)) and g(f(x)).State the domain

Question

1). f(x)=3x+2; g(x)=x-1

2). f(x)=1/x-1; g(x)=√x

3). f(x)=x^2-2; g(x)=√x+1

4). f(x)=1/x+1; g(x)=1/x-1

Answer 1

Elba,

Finding f(g(x)) just means taking the function for g(x) and subbing it in for x in f(x). Same thing for g(f(x)). That should make sense as you look at these solved examples anyway.

Domain is just what can your solutions can be. So, what can x be? In question 1, domain is “Any real number” for both. But in question 2, f(g(x)) has a domain of any real number except 1 (because subbing 1 in for x makes the fraction denominator 0 and you know a fraction can never have 0 for a denominator. 3 has domain of “any real number”. 4 f(g(x)) has domain of any real number except 0 and 4 (g(f(x)) has domain of any real number except 0 for the same reason- you can’t have a denominator = 0.

1) f(g(x)) = 3(x - 1) + 2

g(f(x)) = (3x + 2) - 1

2) f(g(x)) = 1/√x - 1

g(f(x)) = √(1/x - 1)

3) f(g(x)) = (√x + 1)^2 - 2

g(f(x)) = √(x^2 - 2) + 1

4) f(g(x)) = 1/((1/x-1) +1)

g(f(x)) = 1/((1/x +1) - 1)

Hope this helps! :)