If \(x\) is real, the minimum value of \(x^{2}-8x+17\) is
Answer & explanation
Correct answer: option 3
\(\begin{aligned}f^{\prime}(x)&=2x-8\\ f^{\prime}(x)&=0\\ x&=4\\ f^{\prime \prime}&=2>0\end{aligned}\hspace{8cm}\)
Minimum value \(=16-32+17=1\)
Correct answer: option 3