Question

find the inverse of the function. \\( n(x) = sqrt3{x} - 2 \\) write your answer in the form \\( a(bx + c)^3 + d \\), where \\( a, b, c, \\) and \\( d \\) are constants. simplify any fractions. \\( n^{-1}(x) = \\)

Explanation:

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

We start by setting \( y = \sqrt[3]{x} - 2 \). This is the original function \( h(x) \) rewritten with \( y \) instead of \( h(x) \).
\[ y = \sqrt[3]{x} - 2 \]

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

To find the inverse, we swap the roles of \( x \) and \( y \). So we get:
\[ x = \sqrt[3]{y} - 2 \]

Step3: Solve for \( y \)

First, we add 2 to both sides of the equation to isolate the cube - root term:
\[ x + 2=\sqrt[3]{y} \]
Then, to eliminate the cube - root, we cube both sides of the equation. Recall that \( (\sqrt[3]{a})^3=a \) for any real number \( a \). So, cubing both sides gives us:
\[ (x + 2)^3=y \]
Which means \( h^{-1}(x)=(x + 2)^3 \) or expanding \( (x + 2)^3=x^{3}+6x^{2}+12x + 8 \), but in the form \( a(bx + c)^3 + d \), we have \( a = 1 \), \( b = 1 \), \( c = 2 \), \( d = 0 \) since \( (x+2)^3=1\times(1\times x + 2)^3+0 \)

Answer:

\( h^{-1}(x)=(x + 2)^3 \) (or in the form \( 1(x + 2)^3+0 \))