Function Composition and Differentiation

Composition of Two Functions

• Definition of composite functions:

  • Let $u$ be a function defined on an interval $I$.
  • Let $v$ be a function defined on an interval J$.\n - Assume that for all real numbers x imes I$, the image$u(x)$belongs to$J$.
  • The composite function of $u$ followed by $v$, denoted $v \circ u$ (read as "$v$ circle $u$" or "$v$ round $u$"), is the function defined on $I$ by:     $(v \circ u)(x) = v(u(x))$

• Sequential evaluation and domain conditions:

  • The operation proceeds sequentially: input $x$ is evaluated first under $u$ to give intermediate result $u(x)$, which then serves as the input for $v$
  • For $v(u(x))$ to be well-defined for all $x \in I$, the range of $u$ over $I$ must be contained within the domain $J$ of $v$

Decomposition and Composition Analysis

• Example Problem:

  • Let $f$ be the function defined on $\mathbb{R} \setminus \{1\}$ by:     $f(x) = e^{\frac{x}{x-1}}$
  • Decomposition Task: Express $f$ in the form $v \circ u$ by specifying the individual functions $u$ and $v$
  • Recomposition Task: Define the composite function $u \circ v$ and determine its exact domain of definition

• Solution for Part a (Decomposing $f(x)$ into $v \circ u$):

  • Mapping chain:     $x \xrightarrow{u} \frac{x}{x-1} \xrightarrow{v} e^{\frac{x}{x-1}}$
  • Determination of Inner Function $u$:
  • Expression: $u(x) = \frac{x}{x-1}$
  • Domain restriction: The denominator cannot be zero, requiring $x - 1 \neq 0$, which yields $x \neq 1$
  • Domain of $u$: $I = \mathbb{R} \setminus \{1\}$
  • Determination of Outer Function $v$:
  • Expression: $v(x) = e^x$
  • Domain of $v$: $\mathbb{R}$
  • Conclusion for Decomposition:
  • $f = v \circ u$, where $u$ is defined on $\mathbb{R} \setminus \{1\}$ by $u(x) = \frac{x}{x-1}$ and $v$ is defined on $\mathbb{R}$ by $v(x) = e^x$

• Solution for Part b (Defining $g = u \circ v$ and Domain Analysis):

  • Composite mapping chain for $g = u \circ v$:     $x \xrightarrow{v = \exp} e^x \xrightarrow{u} \frac{e^x}{e^x - 1}$
  • Step-by-step domain derivation:
  • Input $x$ must be in the domain of $v(x) = e^x$, which is $\mathbb{R}$
  • Output $v(x) = e^x$ must lie in the domain of $u$, which is $\mathbb{R} \setminus \{1\}$
  • Requirement: $e^x \neq 1$
  • Since $e^0 = 1$, the equation $e^x = 1$ holds if and only if $x = 0$
  • Therefore, the restriction $e^x \neq 1$ requires $x \neq 0$
  • Conclusion for $g = u \circ v$:
  • The composite function $g = u \circ v$ is defined on domain $\mathbb{R} \setminus \{0\}$ with explicit formula:       $g(x) = \frac{e^x}{e^x - 1}$