Question

If \(f:R\to R\) is given by \(f(x)=(5-x^{5})^{1/5}\), then \(f\circ f(x)\) is equal to:

• A. \(x^{1/5}\)
• B. \(x\)
• C. \(x^{5}\)
• D. \(5-x^{5}\)

Hint:

Substitute \(f(x)\) into itself; the 5th power and 5th root cancel.

Answer 1

The Correct Option is B

Step 1: Testing invertibility:
A function of the form \(f(x)=(c-x^{n})^{1/n}\) is always its own inverse, i.e. \(f=f^{-1}\), because solving \(y=(c-x^n)^{1/n}\) for \(x\) gives \(x=(c-y^n)^{1/n}\), the same rule.

Step 2: Applying the self-inverse property:
Since \(f\) is its own inverse, \(f(f(x))=f^{-1}(f(x))=x\) for every \(x\) in the domain.

Final Answer:
\[ \boxed{x} \]