Can someone please explain to me what a combined function is and how it is different from a composite function.
From my understanding a combined function would be $(f+g)(x) = f(x)+g(x)$ and a composite function is $(f\circ g)(x)=f\big(g(x)\big)$
Also what factors affect a combined functions domain?
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$\begingroup$Using the example functions $f(x)= 2x^2+5x+2$ and $g(x)= 2x+1$ whose domains are the real numbers.
A combined function is an algebraic combination of two function: sum, difference, product, quotient. $$\begin{align}(f+g)(x) & = f(x)+g(x) \\ & = (2x^2+5x+2)+(2x+1) \\ & = 2x^2+7x+3 \\[2ex] (f-g)(x) & = f(x)-g(x) \\ & = (2x^2+5x+2)-(2x+1) \\ &= 2x^2+3x+1 \\[2ex] (f\cdot g)(x) & = f(x)\cdot g(x) \\ & = (2x^2+5x+2)\cdot(2x+1)\\ & = 4x^3+12x^2+9x+2 \\[2ex] (f\div g)(x) & = \frac{f(x)}{g(x)} & \big[g(x)\neq 0\big] \\ & = \frac{ (2x^2+5x+2)}{(2x+1)} & \big[2x+1\neq 0\big]\\ &= x+2 & \big[x\neq -1/2\big]\end{align}$$
The domain of a combined function is the intersection of the domains of the functions which are combined. The domain of a quotient function is further restricted to exclude those points in the denominator function whose image is zero. (vis $g(x)\neq 0$).
Note: in particular, that although $x+2$ can be defined at $x=-\tfrac 1 2$, the domain of the quotient function actually excludes this point. Our $(f\div g)(x)$ is discontinuous at the limit point $x=-\tfrac 1 2$ because $g(-\tfrac 1 2)=0$. It's a subtle but important point when finding valid solutions to systems of equations.
A composite function is a composition of functions (the composition operator is $\circ$ ). $$\begin{align}(f\circ g)(x) & = f\big(g(x)\big) & \neq (\;g\circ f\;)(x) \\ & = f(2x+1) \\ & = 2(2x+1)^2+5(2x+1)+2 \\ & = 8x^2+18x+9\end{align}$$
The domain of a composite function is the domain of the outer function whose image intersects the domain of the inner function. (Order is important here, composition is not commutative).
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