Let a function is defined as `f(x)=(a^(x)+a^(-x))/(2), a gt 0`, what is the value of `f(x+y)+f(x-y)`?

Options

A. `f(x)+f(y)`
B. `f(x)f(y)`
C. `2f(x)f(y)`
D. `4f(x)f(y)`

Answer

To solve the problem, we need to find the value of \( f(x+y) + f(x-y) \) given the function \( f(x) = \frac{a^x + a^{-x}}{2} \).

### Step 1: Calculate \( f(x+y) \)

Using the definition of the function:
\[ f(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2} \]

### Step 2: Calculate \( f(x-y) \)

Similarly, we calculate:
\[ f(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2} \]

### Step 3: Add \( f(x+y) \) and \( f(x-y) \)

Now, we add the two results from Step 1 and Step 2:
\[ f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)}}{2} + \frac{a^{x-y} + a^{-(x-y)}}{2} \]

### Step 4: Combine the fractions

Combining the fractions gives:
\[ f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}}{2} \]

### Step 5: Simplify the expression

We can group the terms:
\[ = \frac{(a^{x+y} + a^{x-y}) + (a^{-(x+y)} + a^{-(x-y)})}{2} \]

### Step 6: Factor out common terms

Notice that:
\[ a^{x+y} + a^{x-y} = a^x(a^y + a^{-y}) \quad \text{and} \quad a^{-(x+y)} + a^{-(x-y)} = a^{-x}(a^{-y} + a^y) \]

Thus, we can write:
\[ = \frac{a^x(a^y + a^{-y}) + a^{-x}(a^{-y} + a^y)}{2} \]

### Step 7: Recognize the structure

Since \( a^y + a^{-y} = 2f(y) \):
\[ = \frac{a^x \cdot 2f(y) + a^{-x} \cdot 2f(y)}{2} \]
\[ = f(y)(a^x + a^{-x}) \]

### Step 8: Substitute back to find the final result

Since \( a^x + a^{-x} = 2f(x) \):
\[ = f(y) \cdot 2f(x) \]

Thus, the final result is:
\[ f(x+y) + f(x-y) = f(x)f(y) \]

### Final Answer:
\[ f(x+y) + f(x-y) = f(x)f(y) \]