Inverse Functions and Their Properties

I. Understanding Inverse Functions

• Definition: An inverse function, denoted by $f^{-1}(x)$, "undoes" the action of the original function $f(x)$. If $f(a)=b$, then $f^{-1}(b)=a$.
• Graphical Relationship: If a point $(a, b)$ lies on the graph of $f(x)$, then the point $(b, a)$ lies on the graph of $f^{-1}(x)$. This characteristic means the graph of $f^{-1}(x)$ is a reflection of the graph of $f(x)$ across the line $y=x$.

II. Steps to Find the Inverse Function $(f^{-1}(x))$

1. Replace $f(x)$ with $y$: Begin by setting the function notation equal to $y$. For example, if you have $f(x) = 2x+3$, rewrite it as $y = 2x+3$.
2. Swap $x$ and $y$: This is the crucial conceptual step where the roles of the input and output are interchanged. The equation $y = 2x+3$ becomes $x = 2y+3$.
3. Solve for $y$ in terms of $x$: This is often the most algebraically intensive step, requiring manipulation to isolate $y$. Examples of algebraic operations include:
   • Isolating terms: Move terms not containing $y$ to the other side of the equation.
   • Eliminating roots/powers: If the variable $y$ is inside a root (e.g., cube root), raise both sides to the corresponding power (e.g., to the power of 3) to eliminate it. Conversely, if $y$ is squared or cubed, take the corresponding root to solve for $y$. For instance, to "get rid of a cube," you would cube both sides.
   • Factoring: If $y$ appears in multiple terms on one side of the equation after swapping, factor out $y$.
     • Example Scenario: Consider an equation like $-xy - 3y = -2x - 4$ (after swapping variables).
     • Factor out $y$ from the terms on the left: $y(-x - 3) = -2x - 4$.
   • Division: After factoring, divide both sides by the remaining factor to completely isolate $y$.
     • Continuing Example: y = rac{-2x - 4}{-x - 3}.
4. Replace $y$ with $f^{-1}(x)$: Once $y$ is isolated, the resulting expression is the inverse function. For the example $f(x)=2x+3$, the inverse would be f^{-1}(x) = rac{x-3}{2}.

III. One-to-One Functions: A Prerequisite for Inverses

• Definition: A function $f(x)$ is one-to-one if every output (range element) corresponds to exactly one input (domain element). In other words, if $f(a) = f(b)$, then $a$ must equal $b$. Conversely, if $a <br /> eq b$, then $f(a) <br /> eq f(b)$.
• Horizontal Line Test: Graphically, a function is one-to-one if and only if no horizontal line intersects its graph more than once.
• Significance for Inverses: For the inverse relation to also be a function, the original function must be one-to-one. If a function is not one-to-one, its inverse would not pass the vertical line test and thus would not be a function.
• Example of a function that is NOT one-to-one:
  • Consider the function $f(x) = (x-4)^2$.
  • If we plug in $x=3$, $f(3) = (3-4)^2 = (-1)^2 = 1$.
  • If we plug in $x=5$, $f(5) = (5-4)^2 = (1)^2 = 1$.
  • Since $f(3) = f(5) = 1$, two different inputs (3 and 5) produce the same output (1). Therefore, $f(x) = (x-4)^2$ is not one-to-one over its entire natural domain (-rac{ ext{infinity}}{}, rac{ ext{infinity}}{}). Because of this, it "has no inverse" function over its entire domain.
• Addressing non-one-to-one functions: To obtain an inverse function from a function that is not one-to-one, we must restrict its domain to a specific interval where it is one-to-one (e.g., where it is consistently increasing or consistently decreasing).
  • Continuing example: For $f(x)=(x-4)^2$, we could restrict the domain to $x ext{ extgreater} = 4$ (i.e., the interval $[4, ext{infinity})$) or $x ext{ extless} = 4$ (i.e., the interval $(- ext{infinity}, 4]$). If we choose the restricted domain $[4, ext{infinity})$, then $f(x)$ becomes one-to-one on this interval.

IV. Domain and Range Considerations for Inverse Functions

• Interchangeability: A fundamental property is that the domain of $f(x)$ becomes the range of $f^{-1}(x)$, and conversely, the range of $f(x)$ becomes the domain of $f^{-1}(x)$.
• Implications when taking square roots: When solving for $y$ in the process of finding an inverse, if an operation like taking a square root is involved, it usually introduces a $ext{ extplus/ ext{-}}$ sign (since $ext{ extsqrt{a^2}} = |a|$). To ensure the inverse is a function, we must choose either the positive or negative root. This choice directly corresponds to the restricted domain (and thus the resulting range) of the original function.
  • Example: If we restrict the domain of $f(x) = (x-4)^2$ to $x ext{ extgreater} = 4$.
    • The range of $f(x)$ on this domain is $[0, ext{infinity})$.
    • Therefore, the domain of $f^{-1}(x)$ will be $[0, ext{infinity})$.
    • And the range of $f^{-1}(x)$ will be $[4, ext{infinity})$ (reflecting the restricted domain of $f(x)$).
    • To find $f^{-1}(x)$ for this restricted function:
      1. $y = (x-4)^2$
      2. Swap variables: $x = (y-4)^2$
      3. Take the square root of both sides: $ext{ extsqrt{x}} = ext{ extsqrt{(y-4)^2}}$
      4. This simplifies to $ext{ extsqrt{x}} = |y-4|$.
      5. Since our original domain was $x ext{ extgreater} = 4$ (meaning the original output $y$ was also $y ext{ extgreater} = 4$), it implies that $(y-4)$ must be non-negative. Thus, $|y-4| = y-4$.
      6. So, we have $ext{ extsqrt{x}} = y-4$
      7. Solve for $y$: $y = ext{ extsqrt{x}} + 4$
      8. Therefore, $f^{-1}(x) = ext{ extsqrt{x}} + 4$.
    • The range of this inverse function is indeed $[4, ext{infinity})$, perfectly matching the restricted domain of the original function.