Question

a one-to-one function is given. write an expression for the inverse function.
$f(x) = 7x^3 - 3$
\\(\circ\\) $f^{-1}(x) = \frac{x - 3}{7}$
\\(\circ\\) $f^{-1}(x) = \frac{x + 3}{7}$
\\(\circ\\) $f^{-1}(x) = \sqrt3{\frac{x + 3}{7}}$
\\(\circ\\) $f^{-1}(x) = \left(\frac{x + 3}{7}\
ight)^3$

Explanation:

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

We start with the function \( f(x) = 7x^3 - 3 \). Replace \( f(x) \) with \( y \), so we have \( y = 7x^3 - 3 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = 7y^3 - 3 \).

Step3: Solve for \( y \)

First, add 3 to both sides of the equation: \( x + 3 = 7y^3 \). Then, divide both sides by 7: \( \frac{x + 3}{7} = y^3 \). Finally, take the cube root of both sides to solve for \( y \): \( y = \sqrt[3]{\frac{x + 3}{7}} \). Since \( y \) is the inverse function \( f^{-1}(x) \), we have \( f^{-1}(x) = \sqrt[3]{\frac{x + 3}{7}} \).

Answer:

\( f^{-1}(x) = \sqrt[3]{\frac{x + 3}{7}} \) (the third option)