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For the curve \(y = 9x + \dfrac{1}{x}\), find the coordinates of the turning points
\begin{align}
&\begin{aligned}
y &=9 x+x^{-1} \\
y^{\prime} &=9-x^{-2} \\
&=9-\frac{1}{x^{2}}
\end{aligned}\\
&\text { Let } y^{\prime}=0 \\
&\begin{aligned}
\therefore \quad 9-\frac{1}{x^{2}} &=0 \\
\frac{1}{x^{2}} &=9 \\
\therefore x^{2} &=\frac{1}{9} \\
x &=\pm \frac{1}{3}
\end{aligned}\\
&\text { when } x=\frac{1}{3} \\
&\begin{aligned}
y &=9 \times \frac{1}{3}+3 \\
&=6
\end{aligned} \\
&\text { when } x=-\frac{1}{3} \\
&\begin{aligned}
y&=9 \times-\frac{1}{3}-3 \\
& =-6
\end{aligned}\\
&\therefore \text { Point are: } \\
&\left(\frac{1}{3}, 6\right) \text { and }\left(-\frac{1}{3},-6\right)
\end{align}
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