Find the Inverse of Cube Root Functions

• Domain and range of the given function:
  \[D = (-\infty, +\infty), \quad R = (-\infty, +\infty)\]
• Write the function as an equation:
  \[y = \sqrt[3]{2x - 1}\]
• Cube both sides:
  \[y^3 = 2x - 1\]
• Solve for \(x\):
  \[x = \tfrac{1}{2}(y^3 + 1)\]
• Interchange \(x\) and \(y\):
  \[f^{-1}(x) = \tfrac{1}{2}(x^3 + 1)\]

Conclusion: The domain and range of the inverse function are the same as those of the original function: \[(-\infty, +\infty)\]

• Domain and range:
  \[D = (-\infty, +\infty), \quad R = (-\infty, +\infty)\]
• Write as an equation:
  \[y = \sqrt[3]{\tfrac{x}{3} - 1} - 4\]
• Rearrange:
  \[\sqrt[3]{\tfrac{x}{3} - 1} = y + 4\]
• Cube both sides:
  \[\tfrac{x}{3} - 1 = (y + 4)^3\]
• Solve for \(x\):
  \[x = 3\big((y + 4)^3 + 1\big) = 3(y + 4)^3 + 3\]
• Interchange variables:
  \[f^{-1}(x) = 3(x + 4)^3 + 3\]

Conclusion: The inverse function also has domain and range \((-\infty, +\infty)\).

• Because the domain is restricted to \(x \ge 0\), the function is one-to-one.
• Domain and range of the given function:
  \[D = [0, +\infty), \quad R = [4, +\infty)\]
• Write the function as an equation:
  \[y = \sqrt[3]{4x^2 + 8} + 2\]
• Rearrange and cube both sides:
  \[(y - 2)^3 = 4x^2 + 8\]
• Solve for \(x\):
  \[4x^2 = (y - 2)^3 - 8 \implies x^2 = \tfrac{1}{4}\big((y - 2)^3 - 8\big)\]
• Select the positive root (since \(x \ge 0\)):
  \[x = \sqrt{\tfrac{1}{4}\big((y - 2)^3 - 8\big)} = \tfrac{1}{2}\sqrt{(y - 2)^3 - 8}\]
• Interchange variables:
  \[f^{-1}(x) = \tfrac{1}{2}\sqrt{(x - 2)^3 - 8}\]

Conclusion: The domain and range of the inverse are:
\[D_{f^{-1}} = [4, +\infty), \quad R_{f^{-1}} = [0, +\infty)\]