Verifying Inverse Functions: Algebraic Compositions and Direct Derivation page 33

Fundamentals of Inverse Functions

• Definition of Inverse Functions:

  • Two functions $f$ and $g$ are inverse functions of each other if and only if the composition of $f$ with $g$ and $g$ with $f$ yields the identity function on their respective domains.

  • Formally, $g = f^{-1}$ if both of the following conditions are satisfied:

  • $f(g(x)) = x$ for every $x$ in the domain of $g(x)$.

  • $g(f(x)) = x$ for every $x$ in the domain of $f(x)$.

• Domain and Range Relationships:

  • The domain of $f(x)$ corresponds to the range of $f^{-1}(x)$.

  • The range of $f(x)$ corresponds to the domain of $f^{-1}(x)$.

  • For cubic functions such as $f(x) = (x - 1)^3$ and cube root functions such as $g(x) = \sqrt[3]{x} + 1$, both domain and range encompass all real numbers, denoted as $(-\infty, \infty)$.

Problem Analysis and Given Equations

• Given Functions:

  • Primary function: $f(x) = (x - 1)^3$

  • Proposed candidate inverse function: $g(x) = \sqrt[3]{x} + 1$

• Core Question:

  • Determine whether $g(x)$ is the true inverse of $f(x)$, denoted as $g = f^{-1}$.

• Reference Diagram:

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Verification via Function Composition

• Evaluating First Composition $f(g(x))$:

  • Substitute $g(x)$ into $f(x)$ in place of $x$:

  • $f(g(x)) = f(\sqrt[3]{x} + 1)$

  • Apply the formula for $f(x) = (x - 1)^3$:

  • $f(\sqrt[3]{x} + 1) = ((\sqrt[3]{x} + 1) - 1)^3$

  • Simplify the inner binomial expression:

  • $\sqrt[3]{x} + 1 - 1 = \sqrt[3]{x}$

  • Cube the resulting expression:

  • $(\sqrt[3]{x})^3 = x$

  • Result of first composition:

  • $f(g(x)) = x$ for all real numbers $x$.

• Evaluating Second Composition $g(f(x))$:

  • Substitute $f(x)$ into $g(x)$ in place of $x$:

  • $g(f(x)) = g((x - 1)^3)$

  • Apply the formula for $g(x) = \sqrt[3]{x} + 1$:

  • $g((x - 1)^3) = \sqrt[3]{(x - 1)^3} + 1$

  • Simplify the cube root of the cubic expression:

  • $\sqrt[3]{(x - 1)^3} = x - 1$

  • Add $1$ to the simplified expression:

  • $(x - 1) + 1 = x$

  • Result of second composition:

  • $g(f(x)) = x$ for all real numbers $x$.

Direct Derivation Method

• Alternate Approach: Finding $f^{-1}(x)$ Directly from $f(x)$:

  • Step 1: Replace $f(x)$ with $y$:

  • $y = (x - 1)^3$

  • Step 2: Interchange the variables $x$ and $y$ to set up the inverse equation:

  • $x = (y - 1)^3$

  • Step 3: Solve for $y$ in terms of $x$:

  • Take the cube root of both sides of the equation:

    • $\sqrt[3]{x} = \sqrt[3]{(y - 1)^3}$

    • $\sqrt[3]{x} = y - 1$

  • Isolate $y$ by adding $1$ to both sides:

    • $y = \sqrt[3]{x} + 1$

  • Step 4: Substitute $f^{-1}(x)$ back in for $y$:

  • $f^{-1}(x) = \sqrt[3]{x} + 1$

Conclusion

• Comparison and Verification:

  • The directly derived inverse function $f^{-1}(x) = \sqrt[3]{x} + 1$ matches the given function $g(x) = \sqrt[3]{x} + 1$.

  • Both algebraic compositions $f(g(x)) = x$ and $g(f(x)) = x$ simplify to $x$ across all real numbers.

• Final Answer:

  • Yes, $g = f^{-1}$.