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If the solve the problem

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).

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