Let $I$ be any interval such that $I \cap [-1, 1] = \phi$. Prove that the function $f$ given by $f(x) = x + \frac{1}{x}$ is strictly increasing on $I$.

(N/A) We have,$f(x) = x + \frac{1}{x}$.
Taking the derivative with respect to $x$,we get $f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$.
For the function to be strictly increasing,we require $f'(x) > 0$.
Since $x^2 > 0$ for all $x \neq 0$,the sign of $f'(x)$ depends on the numerator $x^2 - 1$.
$f'(x) > 0 \iff x^2 - 1 > 0 \iff x^2 > 1 \iff |x| > 1$.
This implies $x > 1$ or $x < -1$.
Thus,$f'(x) > 0$ for $x \in (-\infty, -1) \cup (1, \infty)$.
Given that $I$ is an interval such that $I \cap [-1, 1] = \phi$,it follows that $I \subset (-\infty, -1)$ or $I \subset (1, \infty)$.
In both cases,$f'(x) > 0$ for all $x \in I$.
Therefore,the function $f$ is strictly increasing on $I$.