Find the inverse of [tex]$f(n)=2(n+2)^3$[/tex].

A. [tex]f^{-1}(n)=\sqrt[3]{-n-1}[/tex]
B. [tex]f^{-1}(n)=\frac{-4+\sqrt[3]{16n}}{2}[/tex]
C. [tex]f^{-1}(n)=-n^3+1[/tex]
D. [tex]f^{-1}(n)=\sqrt[3]{n}+1[/tex]



Answer :

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.