`f(x)=x^(2)+16`. For what value of k is `f(2k+1)=2f(k)+1` if k is a positive integer?

Answer

To solve the equation \( f(2k + 1) = 2f(k) + 1 \) for the function \( f(x) = x^2 + 16 \), we will follow these steps:

### Step 1: Find \( f(2k + 1) \)
We start by substituting \( 2k + 1 \) into the function \( f(x) \).

\[ f(2k + 1) = (2k + 1)^2 + 16 \]

Expanding \( (2k + 1)^2 \):

\[ (2k + 1)^2 = 4k^2 + 4k + 1 \]

Thus,

\[ f(2k + 1) = 4k^2 + 4k + 1 + 16 = 4k^2 + 4k + 17 \]

### Step 2: Find \( 2f(k) + 1 \)
Next, we calculate \( f(k) \) and then find \( 2f(k) + 1 \).

\[ f(k) = k^2 + 16 \]

Now, multiplying by 2 and adding 1:

\[ 2f(k) = 2(k^2 + 16) = 2k^2 + 32 \]

Thus,

\[ 2f(k) + 1 = 2k^2 + 32 + 1 = 2k^2 + 33 \]

### Step 3: Set the two expressions equal to each other
Now we set \( f(2k + 1) \) equal to \( 2f(k) + 1 \):

\[ 4k^2 + 4k + 17 = 2k^2 + 33 \]

### Step 4: Rearrange the equation
We can rearrange this equation to bring all terms to one side:

\[ 4k^2 + 4k + 17 - 2k^2 - 33 = 0 \]

This simplifies to:

\[ 2k^2 + 4k - 16 = 0 \]

### Step 5: Simplify the equation
Dividing the entire equation by 2 gives:

\[ k^2 + 2k - 8 = 0 \]

### Step 6: Factor the quadratic equation
Now we factor the quadratic equation:

\[ (k - 2)(k + 4) = 0 \]

### Step 7: Solve for \( k \)
Setting each factor to zero gives us:

\[ k - 2 = 0 \quad \Rightarrow \quad k = 2 \]
\[ k + 4 = 0 \quad \Rightarrow \quad k = -4 \]

Since we are looking for the positive integer value of \( k \), we find:

\[ k = 2 \]

### Final Answer
The value of \( k \) is \( \boxed{2} \).

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