Prove that if `f (x)` is periodic function with period T, then the function `f (ax+b)` where `a gt 0,` is periodic with period

Options

A. T/b
B. aT
C. bT
D. T/a

Explanation

To prove that if \( f(x) \) is a periodic function with period \( T \), then the function \( f(ax + b) \) where \( a > 0 \) is periodic with period \( \frac{T}{a} \), we can follow these steps:

### Step 1: Understand the definition of periodicity
A function \( f(x) \) is said to be periodic with period \( T \) if for all \( x \):
\[ f(x + T) = f(x) \]

### Step 2: Define the new function
Let \( g(x) = f(ax + b) \). We need to show that \( g(x) \) is periodic and find its period.

### Step 3: Assume the period of \( g(x) \)
Assume that \( g(x) \) has a period \( K \). This means:
\[ g(x + K) = g(x) \]

### Step 4: Substitute the definition of \( g(x) \)
Using the definition of \( g(x) \):
\[ g(x + K) = f(a(x + K) + b) = f(ax + aK + b) \]
And we know:
\[ g(x) = f(ax + b) \]

### Step 5: Set the two expressions equal
Since \( g(x + K) = g(x) \), we have:
\[ f(ax + aK + b) = f(ax + b) \]

### Step 6: Use the periodicity of \( f(x) \)
Since \( f(x) \) is periodic with period \( T \), we can write:
\[ f(ax + aK + b) = f(ax + b) \implies ax + aK + b = ax + b + nT \]
for some integer \( n \).

### Step 7: Simplify the equation
Subtract \( ax + b \) from both sides:
\[ aK = nT \]

### Step 8: Solve for \( K \)
Now, we can express \( K \) as:
\[ K = \frac{nT}{a} \]

### Step 9: Find the smallest positive period
To find the smallest positive period, we can take \( n = 1 \):
\[ K = \frac{T}{a} \]

### Conclusion
Thus, we have shown that \( g(x) = f(ax + b) \) is periodic with period \( \frac{T}{a} \).