Question 12: Given the function \(y = \frac{{x + 1}}{{{x^2} + 8}}\). Which statement below is correct?

Yes \(y’\left( x \right) = \frac{{ – {x^2} – 2x + 8}}{{{\left( {{x^2} + 8} \right)}^ 2}}}\)

\(y’\left( x \right) = 0 \Leftrightarrow \left[\begin{array}{l}x=–4\\x=2\end{array}\right\)[\begin{array}{l}x=–4\x=2\end{array}\right\)

It is easy to see that \(y’\left( x \right)\) has the same sign as \( – {x^2} – 2x + 8\)

Then x = 2 is the maximum point of the function.

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