If `f(x)=(4^(x))/(4^(x)+2)`, then show that `f(x)+f(1-x)=1`

Explanation

To show that \( f(x) + f(1-x) = 1 \) for the function \( f(x) = \frac{4^x}{4^x + 2} \), we will follow these steps:

### Step 1: Write down the function
We start with the given function:
\[ f(x) = \frac{4^x}{4^x + 2} \]

### Step 2: Find \( f(1-x) \)
Next, we need to calculate \( f(1-x) \):
\[ f(1-x) = \frac{4^{1-x}}{4^{1-x} + 2} \]

### Step 3: Simplify \( f(1-x) \)
We can rewrite \( 4^{1-x} \) as:
\[ 4^{1-x} = \frac{4}{4^x} \]
Thus, we have:
\[ f(1-x) = \frac{\frac{4}{4^x}}{\frac{4}{4^x} + 2} \]

### Step 4: Combine the terms in the denominator
Now, we will combine the terms in the denominator:
\[ f(1-x) = \frac{\frac{4}{4^x}}{\frac{4 + 2 \cdot 4^x}{4^x}} = \frac{4}{4 + 2 \cdot 4^x} \]

### Step 5: Find a common denominator
Now we will add \( f(x) \) and \( f(1-x) \):
\[ f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{4}{4 + 2 \cdot 4^x} \]
The common denominator for these two fractions is \( (4^x + 2)(4 + 2 \cdot 4^x) \).

### Step 6: Add the fractions
Now we will add the two fractions:
\[ f(x) + f(1-x) = \frac{4^x(4 + 2 \cdot 4^x) + 4(4^x + 2)}{(4^x + 2)(4 + 2 \cdot 4^x)} \]

### Step 7: Expand the numerator
Expanding the numerator:
\[ = \frac{4^{x+1} + 2 \cdot 4^{2x} + 4^{x+1} + 8}{(4^x + 2)(4 + 2 \cdot 4^x)} \]
Combine like terms:
\[ = \frac{2 \cdot 4^{2x} + 2 \cdot 4^{x+1} + 8}{(4^x + 2)(4 + 2 \cdot 4^x)} \]

### Step 8: Factor out the numerator
Factoring out 2 from the numerator:
\[ = \frac{2(4^{2x} + 4^{x+1} + 4)}{(4^x + 2)(4 + 2 \cdot 4^x)} \]

### Step 9: Simplify the expression
Notice that \( 4^{2x} + 4^{x+1} + 4 \) can be factored:
\[ = \frac{2(4^x + 2)^2}{(4^x + 2)(4 + 2 \cdot 4^x)} \]
The \( (4^x + 2) \) cancels out:
\[ = \frac{2(4^x + 2)}{4 + 2 \cdot 4^x} \]
This simplifies to:
\[ = 1 \]

### Conclusion
Thus, we have shown that:
\[ f(x) + f(1-x) = 1 \]