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Let I be any interval disjoint from $(-1,1)$. Prove that the function $f$ given by $f(x)=x+\frac{1}{x}$ is strictly increasing on I.
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Verified Answer
$\begin{aligned}
&\mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^2}=\frac{\mathrm{x}^2-1}{\mathrm{x}^2} \text { Now } \mathrm{x} \in \mathrm{I} \Rightarrow \mathrm{x} \notin(-1,1) \\
&\Rightarrow \mathrm{x} \leq-1 \text { or } \mathrm{x} \geq 1 \Rightarrow \mathrm{x}^2 \geq 1 \Rightarrow \mathrm{x}^2-1 \geq 0 \\
&\Rightarrow \frac{\mathrm{x}^2-1}{\mathrm{x}^2} \geq 0 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \text {, Thus } \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \forall \mathrm{x} \in \mathrm{I}
\end{aligned}$
Hence, $\mathrm{f}^{\prime}(\mathrm{x})$ is strictly increasing on I.
&\mathrm{f}^{\prime}(\mathrm{x})=1-\frac{1}{\mathrm{x}^2}=\frac{\mathrm{x}^2-1}{\mathrm{x}^2} \text { Now } \mathrm{x} \in \mathrm{I} \Rightarrow \mathrm{x} \notin(-1,1) \\
&\Rightarrow \mathrm{x} \leq-1 \text { or } \mathrm{x} \geq 1 \Rightarrow \mathrm{x}^2 \geq 1 \Rightarrow \mathrm{x}^2-1 \geq 0 \\
&\Rightarrow \frac{\mathrm{x}^2-1}{\mathrm{x}^2} \geq 0 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \text {, Thus } \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 \forall \mathrm{x} \in \mathrm{I}
\end{aligned}$
Hence, $\mathrm{f}^{\prime}(\mathrm{x})$ is strictly increasing on I.
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