a) The given function is expressed as
f(x) = x^3 + 2
The first step is to find the inverse of the function. We would replace f(x) with y. It becomes
y = x^3 + 2
The next step is to interchange x and y. We have
x = y^3 + 2
Next, we would solve for y. We have
y^3 = x - 2
taking the cube root of both sides of the equation,
![\begin{gathered} y\text{ = }\sqrt[3]{x\text{ - 2}} \\ \text{Changing y to f}^(-1), \\ f^(-1)(x)\text{ = }\sqrt[3]{x\text{ - 2}} \\ \text{Note } \\ a^{(b)/(c)}\text{ = (}\sqrt[c]{a})^b \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fqp7eyebdxotkck3zlf884gy1v5fah7qsf.png)
b) To show that f(f^-1(x)) = x, we would substitute x = the inverse function into the original function. We have
![f(f^(-1)(x))\text{ = (}\sqrt[3]{x\text{ - 2}})^3\text{ + 2 }=(x-2)^{(3)/(3)}+2=x-2+2=x^{}](https://img.qammunity.org/2023/formulas/mathematics/college/1smr6bh06gb0nrhm66pw4vvfacvwiq3ueh.png)
To find f^-1(f(x)), we would substitute x = the original function into the inverse function. We have
![f^(-1)(f(x))=\text{ }\sqrt[3]{x^3\text{ + 2 - 2}}=\text{ }\sqrt[3]{x^3}=x^{(3)/(3)}\text{ = x}](https://img.qammunity.org/2023/formulas/mathematics/college/ks4370tx6tp0xlqtbklg8u3mc338ty1ixz.png)
This is the final part.