Composite Functions

Definition and Fundamental Principles of Composite Functions

• Core Concept:

  • Composite functions combine two or more functions such that the output of one function becomes the input for the next function.

  • Given two functions $f(x)$ and $g(x)$, the composite function $f \circ g$ is evaluated by applying the inside function $g(x)$ first, and then applying the outside function $f(x)$ to the result.

• Formal Definition:

  • Given two functions $f(x)$ and $g(x)$, the composite function $f \circ g$ (read as "$f$ composed with $g$") is defined by:     $(f \circ g)(x) = f(g(x))$

• Sequential Order of Operations:

  • First step: Apply the function $g(x)$ to the initial input $x$.

  • Second step: Apply the function $f(x)$ to the resulting value $g(x)$.

• Domain Rules for Composite Functions:

  • The domain of the composite function $(f \circ g)(x)$ consists of all values of $x$ that satisfy two criteria:

  • The input value $x$ must be contained within the domain of the inside function $g(x)$ (i.e., $x$ must be allowed by $g(x)$).

  • The output value $g(x)$ generated by the inside function must be contained within the domain of the outside function $f(x)$ (i.e., $g(x)$ must be allowed by $f(x)$).

Evaluating Composite Functions at Specific Values

• Base Functions for Numerical Evaluation:

  • Let $f(x) = 5(x) - 1$

  • Let $g(x) = 3(x)^2$

• Evaluating $(f \circ g)(1)$:

  • Definition expression:     $(f \circ g)(1) = f(g(1))$

  • Step 1: Evaluate the inside function $g(1)$ first:     $g(1) = 3(1)^2$     $g(1) = 3(1) = 3$

  • Step 2: Evaluate the outside function $f(3)$ second:     $f(g(1)) = f(3)$     $f(3) = 5(3) - 1$     $f(3) = 15 - 1 = 14$

  • Final value:     $(f \circ g)(1) = 14$

• Evaluating $(g \circ f)(1)$:

  • Definition expression:     $(g \circ f)(1) = g(f(1))$

  • Step 1: Evaluate the inside function $f(1)$ first:     $f(1) = 5(1) - 1$     $f(1) = 5 - 1 = 4$

  • Step 2: Evaluate the outside function $g(4)$ second:     $g(f(1)) = g(4)$     $g(4) = 3(4)^2$     $g(4) = 3(16) = 48$

  • Final value:     $(g \circ f)(1) = 48$

• Evaluating $(f \circ f)(-2)$:

  • Definition expression:     $(f \circ f)(-2) = f(f(-2))$

  • Step 1: Evaluate the inside function $f(-2)$ first:     $f(-2) = 5(-2) - 1$     $f(-2) = -10 - 1 = -11$

  • Step 2: Evaluate the outside function $f(-11)$ second:     $f(f(-2)) = f(-11)$     $f(-11) = 5(-11) - 1$     $f(-11) = -55 - 1 = -56$

  • Final value:     $(f \circ f)(-2) = -56$

• Evaluating $(g \circ g)(-2)$:

  • Definition expression:     $(g \circ g)(-2) = g(g(-2))$

  • Step 1: Evaluate the inside function $g(-2)$ first:     $g(-2) = 3(-2)^2$     $g(-2) = 3(4) = 12$

  • Step 2: Evaluate the outside function $g(12)$ second:     $g(g(-2)) = g(12)$     $g(12) = 3(12)^2$     $g(12) = 3(144) = 432$

  • Final value:     $(g \circ g)(-2) = 432$

Finding Polynomial Composite Functions and Their Domains

• Base Polynomial Functions:

  • Let $f(x) = (x)^2 + 2(x) - 5$

  • Let $g(x) = 3(x) + 1$

• Finding the Expression for $(f \circ g)(x)$:

  • Algebraic formulation:     $(f \circ g)(x) = f(g(x))$     $(f \circ g)(x) = f(3(x) + 1)$

  • Substituting $3(x) + 1$ into $f(x)$:     $f(3(x) + 1) = (3(x) + 1)^2 + 2(3(x) + 1) - 5$

  • Expanding the squared binomial term:     $(3(x) + 1)^2 = 9(x)^2 + 6(x) + 1$

  • Expanding the linear distribution term:     $2(3(x) + 1) = 6(x) + 2$

  • Combining expanded components:     $(f \circ g)(x) = 9(x)^2 + 6(x) + 1 + 6(x) + 2 - 5$

  • Grouping like terms:     $(f \circ g)(x) = 9(x)^2 + (6(x) + 6(x)) + (1 + 2 - 5)$     $(f \circ g)(x) = 9(x)^2 + 12(x) - 2$

• Determining the Domain of $(f \circ g)(x)$:

  • Domain of $f(x)$: All real numbers ($\mathbb{R}$).

  • Domain of $g(x)$: All real numbers ($\mathbb{R}$).

  • Domain of $(f \circ g)(x)$: All real numbers ($\mathbb{R}$).

• Finding the Expression for $(g \circ f)(x)$:

  • Algebraic formulation:     $(g \circ f)(x) = g(f(x))$     $(g \circ f)(x) = g((x)^2 + 2(x) - 5)$

  • Substituting $(x)^2 + 2(x) - 5$ into $g(x)$:     $g((x)^2 + 2(x) - 5) = 3((x)^2 + 2(x) - 5) + 1$

  • Distributing the coefficient $3$:     $3((x)^2 + 2(x) - 5) = 3(x)^2 + 6(x) - 15$

  • Adding the constant term $1$:     $(g \circ f)(x) = 3(x)^2 + 6(x) - 15 + 1$     $(g \circ f)(x) = 3(x)^2 + 6(x) - 14$

• Determining the Domain of $(g \circ f)(x)$:

  • Domain of $f(x)$: All real numbers ($\mathbb{R}$).

  • Domain of $g(x)$: All real numbers ($\mathbb{R}$).

  • Domain of $(g \circ f)(x)$: All real numbers ($\mathbb{R}$).

Determining Domains of Rational Composite Functions

• Base Rational Functions:

  • Let $f(x) = \frac{1}{(x) + 4}$

  • Let $g(x) = \frac{4}{(x) - 3}$

• Finding the Domain of $(f \circ g)(x)$:

  • Step 1: Determine restrictions on the inside function $g(x)$:

  • The denominator of $g(x)$ cannot equal zero:       $(x) - 3 \neq 0$       $x \neq 3$

  • Therefore, $3$ must be excluded from the domain of $(f \circ g)(x)$.

  • Step 2: Determine restrictions on the outside function $f(x)$:

  • The denominator of $f(x)$ cannot equal zero:       $(x) + 4 \neq 0$       $x \neq -4$

  • Step 3: Set $g(x)$ equal to the restricted value of $f(x)$ (which is $-4$):

  • Equation:       $g(x) = -4$       $\frac{4}{(x) - 3} = -4$

  • Multiply both sides by $((x) - 3)$:       $4 = -4((x) - 3)$       $4 = -4(x) + 12$

  • Isolate $x$:       $4 + 4(x) = 12$       $4(x) = 12 - 4$       $4(x) = 8$       $x = 2$

  • Therefore, $2$ must also be excluded from the domain of $(f \circ g)(x)$.

  • Step 4: State the final combined domain:

  • Exclude both $x = 2$ and x = 3$.\n * Domain in set-builder notation:\n      {x \mid x eq 2,\, x eq 3}\n\n* **Finding the Domain of (g \circ f)(x)**:\n * Step 1: Determine restrictions on the inside function f(x):\n * The denominator of f(x) cannot equal zero:\n      (x) + 4 eq 0\n      x eq -4\n * Therefore, -4$must be excluded from the domain of$(g \circ f)(x).\n * Step 2: Determine restrictions on the outside function g(x):\n * The denominator of g(x) cannot equal zero:\n      (x) - 3 eq 0\n      x eq 3\n * Step 3: Set f(x)$equal to the restricted value of$g(x)$(which is$3):\n * Equation:\n      f(x) = 3\n      \frac{1}{(x) + 4} = 3\n * Multiply both sides by ((x) + 4):\n      1 = 3((x) + 4)\n      1 = 3(x) + 12\n * Isolate x:\n      1 - 12 = 3(x)\n      -11 = 3(x)\n      x = -\frac{11}{3}\n * Therefore, -\frac{11}{3}$must also be excluded from the domain of$(g \circ f)(x).\n * Step 4: State the final combined domain:\n * Exclude both x = -4$and$x = -\frac{11}{3}.\n * Domain in set-builder notation:\n      \left{x \mid x eq -4,\, x eq -\frac{11}{3}\right}\n\n# Constructing Rational Composite Functions and Calculating Domains\n\n* **Base Rational Functions**:\n * Let f(x) = \frac{1}{(x) + 3}\n * Let g(x) = \frac{3}{(x) - 2}\n\n* **Finding (f \circ g)(x) and Its Domain**:\n * Step 1: Formulate the composite algebraic expression f(g(x)):\n    (f \circ g)(x) = f\left(\frac{3}{(x) - 2}\right)\n    (f \circ g)(x) = \frac{1}{\frac{3}{(x) - 2} + 3}\n * Step 2: Simplify the denominator by finding a common denominator:\n    \frac{3}{(x) - 2} + 3 = \frac{3}{(x) - 2} + \frac{3((x) - 2)}{(x) - 2}\n    \frac{3 + 3(x) - 6}{(x) - 2} = \frac{3(x) - 3}{(x) - 2}\n * Step 3: Simplify the complex fraction:\n    (f \circ g)(x) = \frac{1}{\frac{3(x) - 3}{(x) - 2}} = \frac{(x) - 2}{3(x) - 3}\n    (f \circ g)(x) = \frac{(x) - 2}{3((x) - 1)}\n * Step 4: Calculate domain exclusions:\n * From the inside function g(x) = \frac{3}{(x) - 2}$, exclude$x = 2$.

  • From the simplified denominator $3((x) - 1)$, exclude x = 1$.\n * Step 5: Final domain specification:\n * Domain in set-builder notation:\n      {x \mid x eq 1,\, x eq 2}\n\n* **Finding (f \circ f)(x) and Its Domain**:\n * Step 1: Formulate the composite algebraic expression f(f(x)):\n    (f \circ f)(x) = f\left(\frac{1}{(x) + 3}\right)\n    (f \circ f)(x) = \frac{1}{\frac{1}{(x) + 3} + 3}\n * Step 2: Simplify the denominator by finding a common denominator:\n    \frac{1}{(x) + 3} + 3 = \frac{1}{(x) + 3} + \frac{3((x) + 3)}{(x) + 3}\n    \frac{1 + 3(x) + 9}{(x) + 3} = \frac{3(x) + 10}{(x) + 3}\n * Step 3: Simplify the complex fraction:\n    (f \circ f)(x) = \frac{1}{\frac{3(x) + 10}{(x) + 3}} = \frac{(x) + 3}{3(x) + 10}\n * Step 4: Calculate domain exclusions:\n * From the inside function f(x) = \frac{1}{(x) + 3}$, exclude$x = -3$.

  • Solve for when $f(x)$ equals the restricted value $-3$:       $\frac{1}{(x) + 3} = -3$       $1 = -3((x) + 3)$       $1 = -3(x) - 9$       $3(x) = -10$       $x = -\frac{10}{3}$

  • Exclude $x = -\frac{10}{3}$.

  • Step 5: Final domain specification:

  • Domain in set-builder notation:       $\left\{x \mid x \neq -\frac{10}{3},\, x \neq -3\right\}$

Proving Equality of Composite Functions

• Problem Statement:

  • Given $f(x) = 2(x) - 5$ and $g(x) = \frac{(x) + 5}{2}$, prove that $(f \circ g)(x) = (g \circ f)(x) = x$ for every $x$ in the domain.

• Proof for $(f \circ g)(x) = x$:

  • Formulate composition:     $(f \circ g)(x) = f(g(x)) = f\left(\frac{(x) + 5}{2}\right)$

  • Substitute into $f(x)$:     $f\left(\frac{(x) + 5}{2}\right) = 2\left(\frac{(x) + 5}{2}\right) - 5$

  • Simplify terms:     $2\left(\frac{(x) + 5}{2}\right) - 5 = ((x) + 5) - 5 = x$

  • Result:     $(f \circ g)(x) = x$

• Proof for $(g \circ f)(x) = x$:

  • Formulate composition:     $(g \circ f)(x) = g(f(x)) = g(2(x) - 5)$

  • Substitute into $g(x)$:     $g(2(x) - 5) = \frac{(2(x) - 5) + 5}{2}$

  • Simplify terms:     $\frac{2(x) - 5 + 5}{2} = \frac{2(x)}{2} = x$

  • Result:     $(g \circ f)(x) = x$

• Conclusion:

  • Both compositions yield identity function outputs:     $(f \circ g)(x) = (g \circ f)(x) = x$

  • Holds true for any real number $x$.