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If \( x \) is real, then the minimum value of \( x^{2}-8 x+17 \) is
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The correct answer is:
\( 11 \)
Given that, $f(x)=x^{2}-8 x+17$
So, $f^{\prime}(x)=2 x-8$
Now, $f^{\prime}(x)=0$
$2 x-8=0 \Rightarrow x=4$
Now, $f^{\prime \prime}(x)=2>0$
Therefore, $f(4)=4^{2}-8(4)+17=1$
So, $f^{\prime}(x)=2 x-8$
Now, $f^{\prime}(x)=0$
$2 x-8=0 \Rightarrow x=4$
Now, $f^{\prime \prime}(x)=2>0$
Therefore, $f(4)=4^{2}-8(4)+17=1$
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