Are the following functions inverses? $$ \begin{array}{l}f(x)=\sqrt[3]{x-4} \ g(x)=(x+4)^{3} \ ext { Yes, they are inverse } \ ext { No, they are not inverses }\end{array} $$

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Are the following functions inverses? $$ \begin{array}{l}f(x)=\sqrt[3]{x-4} \ g(x)=(x+4)^{3} \ 	ext { Yes, they are inverse } \ 	ext { No, they are not inverses }\end{array} $$

UpStudy AI · Accepted Answer

1. We start with the function
   $$
   f(x)=\sqrt[3]{x-4}.
   $$
   Its inverse, denoted $$ f^{-1}(x) $$, should satisfy:
   $$
   f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x.
   $$

2. To find the inverse $$ f^{-1}(x) $$, let
   $$
   y=\sqrt[3]{x-4}.
   $$
   Then, solving for $$ x $$:
   $$
   y^3 = x-4 \quad \Longrightarrow \quad x = y^3 + 4.
   $$
   Interchanging $$ x $$ and $$ y $$, we obtain:
   $$
   f^{-1}(x) = x^3+4.
   $$

3. The given function $$ g(x) $$ is:
   $$
   g(x)=(x+4)^3.
   $$
   Notice that
   $$
   f^{-1}(x)=x^3+4 \quad \neq \quad g(x)=(x+4)^3.
   $$

4. To confirm, we can check the compositions:

- Compute $$ f(g(x)) $$:
     $$
     f(g(x)) = f((x+4)^3)=\sqrt[3]{(x+4)^3-4}.
     $$
     The cube root of $$(x+4)^3$$ is $$x+4$$, but due to the extra $$-4$$, this does not simplify to $$ x $$.

- Compute $$ g(f(x)) $$:
     $$
     g(f(x)) = \left(\sqrt[3]{x-4}+4\right)^3.
     $$
     This also does not simplify to $$ x $$ in general.

5. Since neither composition simplifies to $$ x $$, the functions $$ f(x) $$ and $$ g(x) $$ are not inverses of each other.

$$
\boxed{\text{No, they are not inverses}}
$$
They are not inverses.

Are the following functions inverses? \( \begin{array}{l}f(x)=\sqrt[3]{x-4} \\ g(x)=(x+4)^{3} \\ \text { Yes, they are inverse } \\ \text { No, they are not inverses }\end{array} \)

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