Two functions f and g are inverse functions if fog(x) = x and gof (x) = x for all values of x in the domain of f and g.

For instance, f (x) = 2x and g(x) = x are inverse functions because fog(x) = f (g(x)) = f (x) = 2(x) = x and gof (x) = g(f (x)) = g(2x) = (2x) = x. Similarly, f (x) = x + 1 and g(x) = x - 1 are inverse funcions because fog(x) = f (g(x)) = f (x - 1) = (x - 1) + 1 = x and gof (x) = g(f (x)) = g(x + 1) = (x + 1) - 1 = x. h(x) = 3x - 1 and j(x) = are inverse functions because hoj(x) = h(j(x)) = h() = 3() - 1 = x + 1 - 1 = x and joh(x) = j(h(x)) = j(3x - 1) = = = x.

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

The trick to finding the inverse of a function f (x) is to "undo" all the operations on x in reverse order.

The function f (x) = 2x - 4 has two steps:

1. Multiply by 2.
2. Subtract 4.

Thus, f^-1(x) must have two steps:

1. Add 4.
2. Divide by 2.

Consequently, f^-1(x) = .
We can verify that this is the inverse of f (x):

Example 1: Find the inverse of f (x) = 3(x - 5).

Original function:

1. Subtract 5.
2. Multiply by 3.

New function:

1. Divide by 3.
2. Add 5.

Thus, f^-1(x) = + 5.

Check:

Example 2: Find the inverse of f (x) = , x≥2 (we must restrict the domain because f (x) is undefined for x < 2).

Original function:

1. Subract 2.
2. Take the square root.

New function:

1. Square.
2. Add 2.

Thus, f^-1(x) = x² + 2.

Check:

When we take the inverse of a function, the domain and range switch. In example 2, the domain of f is x≥2 and the range of f is f (x)≥ 0. Thus, the domain of f^-1 is x≥ 0 and the range of f^-1 is f^-1(x)≥2.