If `f(x)=a^(x)`, then the false statement from the following is:

Options

A. `f(-x)*f(x)=1`
B. `f(x+3)-2f(x+2)+f(x+1)=(a-2)^(2)f(x+1)`
C. `f(x+y)=f(x).f(y)`
D. `f(x) div f(y)=f((x)/(y))`

Answer

To determine which statement is false given the function \( f(x) = a^x \), we will analyze each statement one by one.

### Step-by-Step Solution:

1. **Statement 1**: \( f^{-1}(f(-x)) \cdot f(x) = 1 \)

- We know \( f(x) = a^x \) and \( f(-x) = a^{-x} = \frac{1}{a^x} \).
- The inverse function \( f^{-1}(y) \) for \( f(x) = a^x \) is \( f^{-1}(y) = \log_a(y) \).
- Thus, \( f^{-1}(f(-x)) = f^{-1}(a^{-x}) = -x \).
- Therefore, \( f^{-1}(f(-x)) \cdot f(x) = (-x) \cdot a^x \).
- This does not equal 1 for all \( x \), so this statement is **false**.

2. **Statement 2**: \( f(x + 3) - 2f(x + 2) + f(x + 1) = (a - 2)^2 f(x + 1) \)

- Calculate \( f(x + 3) = a^{x + 3} = a^x \cdot a^3 \).
- Calculate \( f(x + 2) = a^{x + 2} = a^x \cdot a^2 \).
- Calculate \( f(x + 1) = a^{x + 1} = a^x \cdot a \).
- Substitute these into the equation:
\[ a^x a^3 - 2(a^x a^2) + (a^x a) = (a - 2)^2 (a^x a) \]
- Factor out \( a^x \):
\[ a^x (a^3 - 2a^2 + a) = (a - 2)^2 (a^x a) \]
- Simplifying gives:
\[ a^3 - 2a^2 + a = (a - 2)^2 a \]
- Expanding \( (a - 2)^2 a \) gives \( a^3 - 4a^2 + 4a \).
- Thus, \( a^3 - 2a^2 + a \neq a^3 - 4a^2 + 4a \), so this statement is **false**.

3. **Statement 3**: \( f(x + y) = f(x) \cdot f(y) \)

- This is true since:
\[ f(x + y) = a^{x+y} = a^x \cdot a^y = f(x) \cdot f(y) \]
- Therefore, this statement is **true**.

4. **Statement 4**: \( \frac{f(x)}{f(y)} = f(xy) \)

- This means \( \frac{a^x}{a^y} = a^{xy} \).
- Simplifying gives \( a^{x-y} = a^{xy} \).
- This is not true for all \( x \) and \( y \), hence this statement is **false**.

### Conclusion:
The false statements are **Statement 1, Statement 2, and Statement 4**.