Function composition and function analysis Knowledge-Based Study

Study notes

1. Definition of Composite Functions

   Function composition is an operation that takes two functions, $f$ and $g$, and produces a new function by applying $f$ to the output of $g$. The composite function is denoted as $f \circ g$ (read as 'f circle g' or 'f composed with g').

   Formally, for any input $x$ in the appropriate domain:

   $(f \circ g)(x) = f(g(x))$

   In this process:

   • The input $x$ is first processed by the inner function $g$ to produce the intermediate value $g(x)$.
   • The intermediate value $g(x)$ then acts as the input to the outer function $f$, producing the final output $f(g(x))$.

   Algebraic Evaluation

   To evaluate a composite function algebraically, substitute the expression of the inner function into every occurrence of the independent variable in the outer function.

   Example: General Form

   Let $f(x) = 2x^2 - 3$ and $g(x) = 3x + 1$.

   • To find $(f \circ g)(x)$, substitute $g(x)$ into $f$:
     $(f \circ g)(x) = f(3x + 1) = 2(3x + 1)^2 - 3 = 2(9x^2 + 6x + 1) - 3 = 18x^2 + 12x - 1$.
   • To find $(g \circ f)(x)$, substitute $f(x)$ into $g$:
     $(g \circ f)(x) = g(2x^2 - 3) = 3(2x^2 - 3) + 1 = 6x^2 - 9 + 1 = 6x^2 - 8$.

   Key Insight: In general, function composition is not commutative; that is, $(f \circ g)(x) \neq (g \circ f)(x)$.

   Determining the Domain

   The domain of $f \circ g$ consists of all real numbers $x$ such that:

   1. $x$ is in the domain of $g$ (the inner function).
   2. $g(x)$ is in the domain of $f$ (the outer function).

   Expressed symbolically, the domain is $\{x \in \text{dom}(g) : g(x) \in \text{dom}(f)\}$.

   Worked Example

   Consider $f(x) = \frac{1}{x - 4}$ and $g(x) = \sqrt{x + 2}$.

   • Step 1: Find the domain of $g$. The radicand must be non-negative: $x + 2 \ge 0$, so $x \ge -2$.
   • Step 2: Find values where $g(x)$ is in the domain of $f$. The denominator of $f$ cannot be zero, so $g(x) \neq 4$, which means $\sqrt{x + 2} \neq 4$. Squaring both sides yields $x + 2 \neq 16$, so $x \neq 14$.
   • Conclusion: Combining these restrictions, the domain of $(f \circ g)(x)$ is $[-2, 14) \cup (14, \infty)$.

   Decomposing Functions

   Decomposition is the reverse process of composition: breaking a complex function $h(x)$ into simpler constituent functions $f$ and $g$ such that $h(x) = f(g(x))$.

   For example, if $h(x) = (5x^3 - 7)^4$, a natural choice is:

   • Inner function: $g(x) = 5x^3 - 7$
   • Outer function: $f(u) = u^4$, yielding $f(g(x)) = (5x^3 - 7)^4$.

   Analytical Properties

   Analyzing composite functions often simplifies determining properties such as parity and inverses:

   • Parity: If $g$ is an even function, then $(f \circ g)(-x) = f(g(-x)) = f(g(x)) = (f \circ g)(x)$, making the composite function even regardless of whether $f$ is even or odd.
   • Inverses: Two functions $f$ and $g$ are inverses if and only if $(f \circ g)(x) = x$ for all $x$ in the domain of $g$, and $(g \circ f)(x) = x$ for all $x$ in the domain of $f$.