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So basically the question is to find the minimum value of the sum

$$f(x)=x+(1/x)$$

for any real number $x$.

I differentiated the function and found the values of $x$ for which $f'(x)=0$ as $-1$ and $1$.

Using the second derivative test I find that maximal is $-1$ and minimal is $+1$. But isn't this counter intuitive? Like, mere substitution tells us $f(-1)$ has a lower value than $f(1)$. Also if it is for any real number, shouldn't the minimal be at $x=-\infty$ and maxima at $x=\infty$?

What am I doing wrong?

Nikunj
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Dangnabbit
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    You're forgetting about the discontinuity at $x=0$. ($1/x$ isn't defined at $0$). Your intuition fails you because anything can happen as you cross an infinite discontinuity. – Michael Burr Apr 04 '16 at 10:46
  • Are you asked to find the global extrema or the local extrema? You've found both (the local extrema are at $x=-1$ and $x=1$), and the global maximum and minimum don't exist. – Michael Burr Apr 04 '16 at 10:48
  • It isn't mentioned which to find. The options involve only integers though, so I guess they're referring to the local minima. Is it like x=1 is the minima of the upper disjointed part and x=(-1) is the maxima of the lower part? – Dangnabbit Apr 04 '16 at 10:51

2 Answers2

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Using the second derivative test I find that maximal is -1 and minimal is +1. But isn't this counter intuitive?

A short look on the graph hints that only local extrema seem to exist:

enter image description here

So what you found are the function values of the local extrema, meaning they are extremal regarding a neighbourhood around them and might not (as it is the case here) extremal regarding the whole domain.

Also if it is for any real number, shouldn't the minimal be at $x=-\infty$ and maxima at $x=\infty$?

The infinite points are not part of the domain, so they are not considered. This function has no (global) minimum or (global) maximum.

mvw
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enter image description here $f(x)=x+\frac 1x$

$x_{\max}=-1, y(-1)=-2; x_{\min}=1, y(1)=2$

Roman83
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