Functions, Inverse Functions, Graphical Properties, and Function Composition

Algebraic Functions and Definitions

• Linear Function Example 1:

  • Function expression: $f(x) = 2x + 5$
  • Function type: Linear function
  • Slope ($m$): $2$
  • Vertical intercept ($y$-intercept, $b$): $5$

• Linear Function Example 2:

  • Function expression: $f(x) = 3x - 7$
  • Function type: Linear function
  • Slope ($m$): $3$
  • Vertical intercept ($y$-intercept, $b$): $-7$

• Quadratic Function Example:

  • Function expression: $f(x) = x^2 + 4$
  • Function type: Quadratic polynomial function
  • Vertex coordinate: $(0, 4)$
  • Direction of opening: Upward

• Rational/Linear Function Example:

  • Function expression: $f(x) = \frac{x + 2}{3}$
  • Linear expanded form: $f(x) = \frac{1}{3}x + \frac{2}{3}$
  • Slope ($m$): $\frac{1}{3}$
  • Vertical intercept ($y$-intercept, $b$): $\frac{2}{3}$

Function Evaluation

• Fundamental Concept of Evaluation:

  • Evaluating a function $f(x)$ at a specific input $x = a$ involves substituting $a$ for every instance of $x$ in the function definition and simplifying the resulting arithmetic expression to obtain $f(a)$.

• Problem 5: Evaluate $f(3)$ for $f(x) = 4x - 9$

  • Target function: $f(x) = 4x - 9$
  • Input value: $x = 3$
  • Substitution: $f(3) = 4(3) - 9$
  • Multiplication step: $f(3) = 12 - 9$
  • Simplified result: $f(3) = 3$

• Problem 6: Evaluate $f(2)$ for $f(x) = x^2 - 5x + 6$

  • Target function: $f(x) = x^2 - 5x + 6$
  • Input value: $x = 2$
  • Substitution: $f(2) = (2)^2 - 5(2) + 6$
  • Squaring and multiplication steps: $f(2) = 4 - 10 + 6$
  • Simplified result: $f(2) = 0$

• Problem 7: Evaluate $f(-4)$ for $f(x) = 2x + 1$

  • Target function: $f(x) = 2x + 1$
  • Input value: $x = -4$
  • Substitution: $f(-4) = 2(-4) + 1$
  • Multiplication step: $f(-4) = -8 + 1$
  • Simplified result: $f(-4) = -7$

Inverse Functions

• Theory and Definition of Inverses:

  • An inverse function $f^{-1}(x)$ undoes the operation of $f(x)$. It maps the range of $f(x)$ back to its domain, satisfying the relations $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.
  • Systematic algebraic procedure for determining an inverse function $f^{-1}(x)$:
  1. Replace $f(x)$ with $y$.
  2. Interchange the variables $x$ and $y$ to represent the inverse mapping.
  3. Solve the equation for $y$ in terms of $x$
  4. Replace $y$ with $f^{-1}(x)$.

• Problem 8: Find $f^{-1}(x)$ for $f(x) = 2x + 3$

  • Step 1 (Replace): $y = 2x + 3$
  • Step 2 (Swap variables): $x = 2y + 3$
  • Step 3 (Isolate $y$ terms): $x - 3 = 2y$
  • Step 4 (Divide by coefficient): $y = \frac{x - 3}{2}$
  • Final inverse expression: $f^{-1}(x) = \frac{x - 3}{2}$

• Problem 9: Find $f^{-1}(x)$ for $f(x) = 5x - 10$

  • Step 1 (Replace): $y = 5x - 10$
  • Step 2 (Swap variables): $x = 5y - 10$
  • Step 3 (Isolate $y$ terms): $x + 10 = 5y$
  • Step 4 (Divide by coefficient): $y = \frac{x + 10}{5}$
  • Alternative simplified form: $f^{-1}(x) = \frac{x}{5} + 2$
  • Final inverse expression: $f^{-1}(x) = \frac{x + 10}{5}$

• Problem 10: Find $f^{-1}(x)$ for $f(x) = \frac{x - 4}{2}$

  • Step 1 (Replace): $y = \frac{x - 4}{2}$
  • Step 2 (Swap variables): $x = \frac{y - 4}{2}$
  • Step 3 (Clear fraction): $2x = y - 4$
  • Step 4 (Isolate $y$): $y = 2x + 4$
  • Final inverse expression: $f^{-1}(x) = 2x + 4$

Graphical Analysis of Functions

• Problem 11: Domain and Range Determination:

  • Domain: The complete set of all possible input values ($x$-values) on the horizontal axis for which the relation or function is defined.
  • Method: Observe the graph from far left to far right to identify minimum and maximum horizontal bounds.
  • Range: The complete set of all possible output values ($y$-values) on the vertical axis produced by the function.
  • Method: Observe the graph from bottom to top to identify minimum and maximum vertical bounds.

• Problem 12: Intercept Identification:

  • $x$-intercept: The point(s) where the graph crosses or touches the horizontal $x$-axis.
  • Algebraic evaluation: Set $y = 0$ or $f(x) = 0$ and solve for $x$. Coordinates are expressed as $(x, 0)$.
  • $y$-intercept: The point where the graph crosses the vertical $y$-axis.
  • Algebraic evaluation: Set $x = 0$ and solve for $y$ or compute $f(0)$. Coordinates are expressed as $(0, y)$.

• Problem 13: Vertical Line Test (VLT):

  • Purpose: Determines whether a visual graph represents a valid mathematical function.
  • Rule: A graph represents a function if and only if no vertical line intersects the graph at more than one point.
  • Rationale: If any vertical line passes through two or more points on the graph, a single input value $x$ corresponds to multiple output values $y$, violating the unique output rule of a function.

• Problem 14: Horizontal Line Test (HLT):

  • Purpose: Determines whether a given function has an inverse that is also a function (i.e., whether the function is one-to-one / injective).
  • Rule: A function has a valid inverse function $f^{-1}(x)$ if and only if no horizontal line intersects the graph of $f(x)$ at more than one point.
  • Rationale: If a horizontal line intersects the graph in two or more locations, multiple distinct input values $x$ produce the exact same output value $y$, causing the inverse relation to map one input to multiple outputs.

Function Composition

• Definition of Composite Functions:

  • Function composition is an operation where the output of an inner function serves directly as the input to an outer function.
  • Notation: $(f \circ g)(x) = f(g(x))$ represents substituting the expression $g(x)$ into the variable $x$ of $f(x)$.

• Problem 15: Find $f(g(x))$ given $f(x) = 2x + 1$ and $g(x) = x - 3$

  • Outer function: $f(x) = 2x + 1$
  • Inner function: $g(x) = x - 3$
  • Substitution step: $f(g(x)) = f(x - 3)$
  • Applying definition of $f$: $2(x - 3) + 1$
  • Distributive expansion: $2x - 6 + 1$
  • Simplified composite function: $f(g(x)) = 2x - 5$

• Problem 16: Find $g(f(x))$ given $f(x) = x^2$ and $g(x) = x - 2$

  • Outer function: $g(x) = x - 2$
  • Inner function: $f(x) = x^2$
  • Substitution step: $g(f(x)) = g(x^2)$
  • Applying definition of $g$: $x^2 - 2$
  • Simplified composite function: $g(f(x)) = x^2 - 2$