Algebra 2
Other functions (advanced)
Cube and cube root functions
20 practice questions
0 video lessons
Theory + worked examples
Cube and Cube Root Functions
Common Core Algebra 2 • Standard F-IF.7b • Other Functions
Cube and Cube Root Functions is a topic in Other Functions in the Common Core State Standards. It is aligned to Standard F-IF.7b, which requires students to graph cube and cube root functions and analyze their key features.
The cube \(x^3\) and cube root \(\sqrt[3]{x}\) are inverses, both defined for all reals, odd, and passing through the origin.
Theory
The cube and cube root functions are inverses:
- \(x^3\): domain and range all reals; odd; through the origin.
- \(\sqrt[3]{x}\): undoes the cube; also all reals.
- Both transform as \((x-h)^3+k\) and \(\sqrt[3]{x-h}+k\).
Unlike square roots, cube roots accept negative numbers.
\(x^3\) and \(\sqrt[3]{x}\) are inverses through the origin.
Cube and cube root features.
Inverse operations:
\[x^3=a\ \Rightarrow\ x=\sqrt[3]{a},\qquad \sqrt[3]{x}=a\ \Rightarrow\ x=a^3\]
Cube roots of negatives are real: \(\sqrt[3]{-8}=-2\).
How to solve
- To undo a cube, take the cube root.
- To undo a cube root, cube both sides.
- Keep the sign — odd powers preserve it.
- For transforms, read \((h,k)\).
Example 1 — Solve a cubic
Solve \(x^3=27\).
Solution
Take the cube root.
| \(x\) | \(=\) | \(\sqrt[3]{27}=3\) |
Example 2 — Solve a cube root
Solve \(\sqrt[3]{x}=2\).
Solution
Cube both sides.
| \(x\) | \(=\) | \(2^3=8\) |
Example 3 — Negative cube
Solve \(x^3=-8\).
Solution
An odd power keeps the sign.
| \(x\) | \(=\) | \(\sqrt[3]{-8}=-2\) |
Example 4 — Transformed cubic
Describe \(f(x)=(x-1)^3+2\).
Solution
Shift the parent right \(1\), up \(2\).
| \(\text{center}\) | \(=\) | \((1,2)\) |
Common pitfalls
Cube roots of negatives are real, unlike square roots.
An odd power keeps the sign of the input.
No extraneous solutions from cubing — it is reversible.
Frequently asked questions
What is the domain of \(x^3\)?
All real numbers.
Can you take the cube root of a negative?
Yes — \(\sqrt[3]{-8}=-2\).
How do you solve \(x^3=27\)?
Take the cube root: \(x=3\).
Are the cube and cube root inverses?
Yes — each undoes the other.
More in Other functions (advanced)