The task is to find the inverse function of the given functions in the image. Let me solve each of the problems step by step.

Steps to find the inverse function:

1. Replace $f(x)$ with $y$.
2. Solve for $x$ in terms of $y$.
3. Switch $x$ and $y$ to get the inverse function, $f^{-1}(x)$.

1) $f(x) = 3x + 1$

Step 1: Replace $f(x)$ with $y$: $y = 3x + 1$

Step 2: Solve for $x$: $y - 1 = 3x \quad \Rightarrow \quad x = \frac{y - 1}{3}$

Step 3: Switch $x$ and $y$: $f^{-1}(x) = \frac{x - 1}{3}$

2) $f(x) = x^3 - 2$

Step 1: Replace $f(x)$ with $y$: $y = x^3 - 2$

Step 2: Solve for $x$: $y + 2 = x^3 \quad \Rightarrow \quad x = \sqrt[3]{y + 2}$

Step 3: Switch $x$ and $y$: $f^{-1}(x) = \sqrt[3]{x + 2}$

3) $f(x) = 2x + 1$

Step 1: Replace $f(x)$ with $y$: $y = 2x + 1$

Step 2: Solve for $x$: $y - 1 = 2x \quad \Rightarrow \quad x = \frac{y - 1}{2}$

Step 3: Switch $x$ and $y$: $f^{-1}(x) = \frac{x - 1}{2}$

4) $f(x) = x^2 + 4$

Note: Since this is a quadratic, it has no inverse unless we restrict the domain. I'll proceed assuming the domain is restricted (i.e., $x \geq 0$).

Step 1: Replace $f(x)$ with $y$: $y = x^2 + 4$

Step 2: Solve for $x$: $y - 4 = x^2 \quad \Rightarrow \quad x = \sqrt{y - 4}$

Step 3: Switch $x$ and $y$: $f^{-1}(x) = \sqrt{x - 4}$

5) $f(x) = 2x + 6$

Step 1: Replace $f(x)$ with $y$: $y = 2x + 6$

Step 2: Solve for $x$: $y - 6 = 2x \quad \Rightarrow \quad x = \frac{y - 6}{2}$

Step 3: Switch $x$ and $y$: $f^{-1}(x) = \frac{x - 6}{2}$

Would you like further explanation or details on any of these steps?

Related Questions:

1. What is the general method to determine if a function has an inverse?
2. Can all quadratic functions have inverses? Why or why not?
3. How can domain restrictions affect the existence of an inverse function?
4. What is the significance of switching $x$ and $y$ in the process of finding the inverse?
5. How do transformations like shifts and stretches affect the inverse of a function?

Tip:

Always check if the function is one-to-one before finding its inverse. Functions that aren't one-to-one will need domain restrictions to have an inverse.