Given $f(x) = x + \frac{1}{x}$. Graphically, the function looks like this: https://www.wolframalpha.com/input/?i=plot+y+%3D+x+%2B+1%2Fx.
I have taken the first derivative and found that $f'(x)=1 – \frac{1}{x^{2}}$.
I've also found the turning points of the function by setting $f'(x)=0$. They are $x=\pm 1$.
I've also played with the signs of the first derivative, that is, figuring out when it's positive or negative.
So are the intervals $(-\infty, -1) \cup (1, \infty)$ on which $f$ is increasing and $(-1, 0) \cup (0, 1)$ on which $f$ is decreasing?
Best Answer
You did not specify open intervals.
$f$ is increasing on $(-\infty,-1]$ and on $[1,\infty).$
$f$ is decreasing on [-1,0) and on (0,1].
A union of a pair of intervals is not necessarily an interval. $(-\infty,-1)\cup (1,\infty)$ is not an interval.
$f$ is NOT decreasing on the (non-interval ) set $C=(-1,0)\cup (0,1)$ because $-1/2$ and $+1/2$ belong to $C$ with $-1/2<+1/2$ and $f(-1/2)<f(+1/2).$