The inverse of the function f(x) is f^-1(x).

To prove that f(x) is invertible, we'll have to prove first that f(x) is bijective.

To prove that f(x) is bijective, we'll have to prove that is one-to-one and on-to function.

1) One-to-one function.

We'll suppose that f(x1) = f(x2)

We'll substitute f(x1) and f(x2) by their expressions:

3x1 + 1= 3x2 + 1

We'll eliminate like terms:

3x1 = 3x2

We'll divide by 3:

x1 = x2

A function is one-to-one if and only if for x1 = x2 => f(x1) = f(x2).

2) On-to function:

For a real y, we'll have to prove that it exists a real x.

y = 3x + 1

We'll isolate x to one side. For this reason, we'll add -1 both side:

y - 1 = 3x

We'll use the symmetric property:

3x = y - 1

We'll divide by 3:

x = (y - 1)/3

x is a real number.

From 1) and 2) we conclude that f(x) is bijective.

If f(x) is bijective => f(x) is invertible.

**f^-1(x) = (x - 1)/3**

To find the inverse of f(x) = 3x +1

f(x) = 3x+1

Subtract 1 from both sides:

f(x) -1 = 3x.

Divide both sides by 3 :

(f(x)-1)/3 = x.

Now swap the f(x) and x to get the inverse.

Therefore f(x) = (x -1 )/3 inverse of f(x) = 3x+1