Answer
$(f \circ g)(x)+h(x) = f(g(x)) + h(x) = 32$
Work Step by Step
In this problem, we are asked to add a composite function to another function. We want to evaluate the composite function first, so we will take $(f \circ g)(x)$ because it is a multiplication operation, which takes precedence over addition, according to order of operations. When we take $(f \circ g)(x)$, this means that we are evaluating the function $f(g(x))$. Therefore, every time we see $x$ in the function $f(x)$, we will plug in the function $g(x)$.
Let us go ahead and set up the problem, separating the functions with parentheses:
$f(g(x)) + h(x) = [4(\frac{1}{2}x + 7)] + (-2x + 4)$
Use distributive property to multiply the coefficient with each term of the binomial in the composite function first:
$f(g(x)) + h(x) = [(4)(\frac{1}{2}x) + (4)(7)] + (-2x + 4)$
Multiply to simplify:
$f(g(x)) + h(x) = (\frac{4}{2}x + 28) + (-2x + 4)$
Divide numerator and denominator by their greatest common factor, which is $2$, in this case:
$f(g(x)) + h(x) = (2x + 28) + (-2x + 4)$
Now, we can get rid of the fractions, but pay attention to operations in front of the binomial:
$f(g(x)) + h(x) = 2x + 28 - 2x + 4$
Group like terms:
$f(g(x)) + h(x) = (2x - 2x) + (28 + 4)$
Combine like terms:
$(f \circ g)(x) = f(g(x)) + h(x) = 32$
