The independent variable is sometimes x, that is, f(x). When the independent variable becomes the formula φ(x) containing X, that is, f(x) becomes f[φ(x)].

In essence, this is a question of the definition and value range of composite functions.

1. When φ(x) is a linear function and φ (x) = x+b, then f [φ (x)] = f (x+b).

We can use the image transformation of left and right translation to explain the transformation of their definition domain and value domain:

Translating left and right may change the domain, but it will never change the range.

For example, f(x)= 1/x, the domain x≠0, and the range y≠0.

F (x-1) =1(x-1), the domain x≠ 1, and the range y≠0.

2. When φ(x) is a linear function and φ (x) = 2x+b, then f [φ (x)] = f (2x+b).

The situation is more complicated.

We can't just use the image transformation of left and right translation to explain the transformation of their definition domain and value domain.

For example, f(x)= 1/x, the domain x≠0, and the range y≠0.

F (2x-1) =1(2x-1), the domain x≠ 1/2, and the range y≠0.

Another example is f (x) = sinx, where x is an acute angle, the domain x∈(0, π/2), and the range (0, 1).

f(2x-π/2)= sin(2x-π/2)=-sin(π/2-2x)=-cos2x，domain (π/4，π/2) range (0， 1)。

3. When φ(x) is a logarithmic function and φ (x) = lnx, then f [φ (x)] = f (lnx).

The situation is more complicated.

For example, f (x) = e x, domain r, and range y>0.

F (lnx) = e (lnx) = x, domain x>0, range y>0.

4. However, there are some general rules:

Given the domain (a, b) of f(x), find the domain of f[φ(x)].

Is from inequality a.

Given the domain (a, b) of f[φ(x)], find the domain of f(x).

Is to find the range of the function φ(x) on (a, b).