# f(x)=2x-4, g(x)=x squared + 5x

What is F o G and Domain

Okay, we are given f and g as functions of x. To compose a function means to use one function, g in this case, as the argument of the other, f in this case. Thus we want f o g = f(g(x))

With f(x) = 2x-4 and g(x) = x^2 + 5x, f o g becomes f(x^2 +5x)= 2(x^2 +5x) - 4.

What we do is substitue the equation for g into each place the argument of f occurs. I'll let you determine the domain.

Incidentally, it might be a little easier to grasp composite functions if you don't use the same variable in both functions. Example: suppose f= 2x-4, as you have above and g is a function of t, say g(t)=3t-1. Then in this case f o g = f(g(t)). To evaluate this, we put g(t) in place of the x in the equation 2x-4, so f o g = 2g(t)-4. We now substitute the equation for g(t) to get f o g = 2(3t-1)-4, which we would simplify to f(g(t))=6(t-1) for this example.

11 years ago

8 months ago

## To find the domain of the composite function f o g, we need to consider the domains of both f and g, and any restrictions imposed by the composition.

The function g(x) = x^2 + 5x is a polynomial function, and polynomials are defined for all real numbers. Therefore, the domain of g is the set of all real numbers.

The function f(x) = 2x - 4 is a linear function, and linear functions are defined for all real numbers. Therefore, the domain of f is also the set of all real numbers.

For the composite function f o g = f(g(x)), we need to make sure that the input values of x are valid for both g and f. In other words, we need to ensure that the output of g(x) lies within the domain of f.

Let's substitute g(x) = x^2 + 5x into f(x) = 2x - 4:

f(g(x)) = f(x^2 + 5x)

Now, let's simplify:

f(x^2 + 5x) = 2(x^2 + 5x) - 4 = 2x^2 + 10x - 4

The resulting function, 2x^2 + 10x - 4, is a quadratic function. Quadratic functions are defined for all real numbers, so the domain of f o g is also the set of all real numbers.

Therefore, the domain of f o g is the set of all real numbers.