# Composite function problem

• Dec 3rd 2010, 05:55 AM
Real9999
Composite function problem
I am struggling in understanding the logic behind obtaining the solution of the following problem.

"Consider the two real functions g(x) = x^(-1) and h(y) = y^2+1. What is the composite function? "

• Dec 3rd 2010, 06:47 AM
Quote:

Originally Posted by Real9999
I am struggling in understanding the logic behind obtaining the solution of the following problem.

"Consider the two real functions g(x) = x^(-1) and h(y) = y^2+1. What is the composite function? "

Are you working out the composite or inverse composite function ?
• Dec 3rd 2010, 06:58 AM
DrSteve
There are two composite functions h(g(x)) and g(h(y))

You compose two functions just by substituting one in for the variable in the other.

For example, h(g(x)) = h(x^(-1)) = (x^(-1))^2+1 = x^(-2) + 1 = 1/x^2 + 1 = (1 +x^2)/x^2
• Dec 3rd 2010, 08:14 AM
wonderboy1953
Quote:

Originally Posted by Real9999
I am struggling in understanding the logic behind obtaining the solution of the following problem.

"Consider the two real functions g(x) = x^(-1) and h(y) = y^2+1. What is the composite function? "

Would it help you to understand that the first step would be to take h(y) = y^2+1 and substitute x to get h(x) = x^2 + 1?
• Dec 3rd 2010, 08:33 AM
DrSteve
I think that this question is poorly worded. Based on the answer, you want to compose g with the inverse of h. I have several issues with this:

1) It doesn't state in the problem that y is necessarily the dependent variable.
2) The inverse of h is not well defined. In order for the inverse to be well defined the domain of h needs to be restricted, but in this problem there is no restriction on the domain given.

In any case, to "solve" this problem you need to first find "the" inverse of h (it's not unique), and then plug this into g.

x = y^2 + 1 implies x - 1 = y^2 implies y = sqrt(x - 1) and y = -sqrt(x - 1).

Now substitute into g.

Again, for some reason the solution indicates to just substitute the first one into g, but that is not really correct.