Question

find the inverse of the function.
$f(x) = x^3 + 5$
write your answer in the form $a \cdot \sqrt3{bx + c} + d$, where $a$, $b$, $c$, and $d$ are constants. simplify any fractions.
$f^{-1}(x) = \square$

Explanation:

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

We start with the function \( f(x) = x^3 + 5 \). Replace \( f(x) \) with \( y \), so we have \( y = x^3 + 5 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = y^3 + 5 \).

Step3: Solve for \( y \)

First, subtract 5 from both sides of the equation: \( x - 5 = y^3 \). Then, take the cube root of both sides to solve for \( y \). The cube root of \( y^3 \) is \( y \), and the cube root of \( x - 5 \) is \( \sqrt[3]{x - 5} \). So, \( y=\sqrt[3]{x - 5} \), which can be written in the form \( a\cdot\sqrt[3]{bx + c}+d \) as \( 1\cdot\sqrt[3]{1\cdot x+(- 5)}+0 \).

Step4: Replace \( y \) with \( f^{-1}(x) \)

Now, replace \( y \) with \( f^{-1}(x) \) to get the inverse function. So, \( f^{-1}(x)=\sqrt[3]{x - 5} \) or in the required form \( 1\cdot\sqrt[3]{1x-5}+0 \).

Answer:

\( \sqrt[3]{x - 5} \) (or in the form \( 1\cdot\sqrt[3]{1x - 5}+0 \))