Question

cubic & cube root functions as inverses quick check
establish which of the following options is the inverse of $f(x) = x^3 + 3$. (1 point)
\\(\circ\\) $f^{-1}(x) = \sqrt3{x + 3}$
\\(\circ\\) $f^{-1}(x) = \sqrt3{x} + 3$
\\(\circ\\) $f^{-1}(x) = \sqrt3{x - 3}$
\\(\circ\\) $f^{-1}(x) = \sqrt3{x} - 3$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

Given \( f(x) = x^3 + 3 \), we write \( y = x^3 + 3 \).

Step2: Swap \( x \) and \( y \)

Interchange \( x \) and \( y \) to get \( x = y^3 + 3 \).

Step3: Solve for \( y \)

Subtract 3 from both sides: \( x - 3 = y^3 \).
Take the cube root of both sides: \( y = \sqrt[3]{x - 3} \).
Replace \( y \) with \( f^{-1}(x) \), so \( f^{-1}(x) = \sqrt[3]{x - 3} \).

Answer:

\( f^{-1}(x) = \sqrt[3]{x - 3} \) (the third option, e.g., if options are labeled as 1,2,3,4: 3. \( f^{-1}(x)=\sqrt[3]{x - 3} \))