What are extreme of function y=x +(1/x)? how you determine extreme?

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degeneratecircle's profile pic

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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, when `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.

sciencesolve's profile pic

Posted on

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.`

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