Question

for the function $f(x) = \sqrt3{x - 6}$, find $f^{-1}(x)$.
answer
$\circ$ $f^{-1}(x) = (x - 6)^3$
$\circ$ $f^{-1}(x) = (x + 6)^3$
$\circ$ $f^{-1}(x) = \sqrt3{x} + 6$
$\circ$ $f^{-1}(x) = x^3 + 6$
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Explanation:

Step1: Let \( y = f(x) \)

Set \( y = \sqrt[3]{x - 6} \).

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

We get \( x = \sqrt[3]{y - 6} \).

Step3: Solve for \( y \)

Cube both sides to eliminate the cube root: \( x^3 = y - 6 \). Then add 6 to both sides: \( y = x^3 + 6 \). So \( f^{-1}(x)=x^3 + 6 \).

Answer:

\( f^{-1}(x) = x^{3} + 6 \) (the fourth option: \( f^{-1}(x) = x^{3} + 6 \))