If x is real, find the minimum value of `x^(2)-8x+ 17`

Answer

To find the minimum value of the function \( f(x) = x^2 - 8x + 17 \), we can follow these steps:

### Step 1: Identify the function
We start with the function:
\[ f(x) = x^2 - 8x + 17 \]

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^2 - 8x + 17) = 2x - 8 \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 2x - 8 = 0 \]

### Step 4: Solve for \( x \)
Solving for \( x \):
\[ 2x = 8 \implies x = 4 \]

### Step 5: Determine if it is a minimum or maximum
To confirm whether this critical point is a minimum or maximum, we check the second derivative:
\[ f''(x) = \frac{d^2}{dx^2}(x^2 - 8x + 17) = 2 \]
Since \( f''(x) = 2 > 0 \), the function is concave up, indicating that \( x = 4 \) is a minimum point.

### Step 6: Find the minimum value
Now, we substitute \( x = 4 \) back into the original function to find the minimum value:
\[ f(4) = 4^2 - 8 \cdot 4 + 17 \]
Calculating this:
\[ f(4) = 16 - 32 + 17 = 1 \]

### Conclusion
Thus, the minimum value of the function \( f(x) = x^2 - 8x + 17 \) is:
\[ \boxed{1} \]