If you want a solution that doesn't use calculus (maybe this is a problem from an algebra 2 class, for example), you can first note that

`f(x)=x+1/x=-(-x+1/(-x))=-f(-x),`

so `f` is an odd function. This means that we only need to focus on positive values of `x` and then we can use symmetry to carry our results over to negative values of `x.` *So assume that* `x>0.` In that case, since squares are always non-negative, we have

`0<=(sqrt(x)-1/(sqrt(x)))^2=x+1/x-2,` so

`x+1/x>=2,` and `2` is a *possible* (local) minimum of `x+1/x.` To see that it is indeed a minimum, just note that equality holds above when `sqrt(x)-1/sqrt(x)=0,` which occurs when `x=1.`

**Thus for `x>0,` the minimum value of `x+1/x` is equal to 2, and occurs when** `x=1.`

**By symmetry, whenÂ **`x<0,` ** the maximum value of `x+1/x` is `-2`, and occurs when** `x=-1.`

It is clear that there is no maximum value of `x+1/x` when `x>0` (just make `x` bigger and `x+1/x` gets bigger if `x>1` ), and by symmetry there is no minimum value if `x<0,` so these must be the only extreme values. Here's the graph.

You may use the following theorem that states that if the function has an extreme value at a point `x = c` and if there exists `f'(x)` , then `f'(c) = 0` .

Hence, reasoning by analogy, you need to find if the equation `f'(x) = 0` has any solutions, such that:

`f'(x) = 1 - 1/x^2 => f'(x) = (x^2 - 1)/(x^2)`

Evaluating `f'(x) = 0` yields:

`f'(x) = 0 => (x^2 - 1)/(x^2) = 0 => x^2 - 1 = 0 => x^2 = 1 => x_(1,2) = +-1`

Hence, the critical values of the function occur at `x = +-1.`

You may evaluate the extremes of the function evaluating the function at `x = -1` and `x = 1` , such that:

`f(-1) = -1 + 1/(-1) = -2`

`f(1) = 1 + 1/1 = 2`

**Hence, evaluating the extremes of the given function, yields `x = -1, y = -2` and **`x = 1, y = 2.`