To find the inverse of the function \( f(n) = 2(n+2)^3 \), we need to follow a step-by-step algebraic procedure. Here's how we can do this:
1. Write the function in terms of \( y \):
\( y = 2(n + 2)^3 \)
2. Switch \( y \) and \( n \) to find the inverse function:
\( n = 2(y + 2)^3 \)
However, this approach should instead provide:
\( x = 2(n + 2)^3 \)
Here we are switching \( f(n) \) with the other variable to find the function's inverse, essentially treating \( x = f(n) \).
3. Solve for \( n \) in terms of \( x \):
\( x = 2(n + 2)^3 \)
Divide both sides by 2:
\( \frac{x}{2} = (n + 2)^3 \)
Take the cube root of both sides:
\( \sqrt[3]{\frac{x}{2}} = n + 2 \)
Subtract 2 from both sides:
\( n = \sqrt[3]{\frac{x}{2}} - 2 \)
Hence, the algebraic inverse of the function \( f(n) = 2(n + 2)^3 \) is:
[tex]\[ f^{-1}(x) = \sqrt[3]{\frac{x}{2}} - 2 \][/tex]
Now, let's examine the results that emerge when solving this equation. The inverse function in the answers presented includes:
1. \( \frac{2^{2/3}x^{1/3}}{2} - 2 \)
2. \( \frac{2^{2/3}x^{1/3}(-1/2 - \sqrt{3}i/2)}{2} - 2 \)
3. \( \frac{2^{2/3}x^{1/3}(-1/2 + \sqrt{3}i/2)}{2} - 2 \)
So, the correct inverse function in standard form for real numbers often simplifies as the first term:
[tex]\[ f^{-1}(x) = \frac{\sqrt[3]{x}}{\sqrt[3]{2}} - 2 \][/tex]
or equivalently,
[tex]\[ f^{-1}(x) = \sqrt[3]{\frac{x}{2}} - 2 \][/tex]
Thus, we have derived the proper form of the inverse function from the given scenario.
