Question:
If $x$ is real, the minimum value $x^{2}-8 x+17$ is
(a) $-1$
(b) 0
(c) 1
(d) 2
Solution:
Let $f(x)=x^{2}-8 x+17$
Differentiating both sides with respect to $x$, we get
$f^{\prime}(x)=2 x-8$
$f^{\prime}(x)=0$
$\Rightarrow 2 x-8=0$
$\Rightarrow x=4$
Now,
$f^{\prime \prime}(x)=2>0$
So, x = 4 is the point of local minimum of f(x).
$\therefore$ Minimum value of $f(x)=f(4)=(4)^{2}-8 \times 4+17=16-32+17=1$
Thus, the minimum value of $x^{2}-8 x+17$ is 1 .
Hence, the correct answer is option (c).