If x is real, then the minimum value of x^2 - 8x + 17 is

The correct answer is Option (3) → $1$ ##

Let $f(x) = x^2 - 8x + 17$

$∴f'(x) = 2x - 8$

So, $f'(x) = 0$ gives $x = 4$

Now, $f''(x) = 2 > 0, \forall x$

So, $x = 4$ is the point of local minima.

$∴$ Minimum value of $f(x)$ at $x = 4$,

$f(4) = 4 \times 4 - 8 \times 4 + 17 = 1$