# Homework Help: Maximum and minimum of a set

1. Apr 7, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! It's kind of the first time that I try to solve this type of problem, so I'd like to see what you guys think about it and if I am heading in the right direction:

Examine if the set {(x2 + x + 1)/(x2 + 1) : x∈ℝ} has a maximum and/or a minimum, and calculate those extrema if so.

2. Relevant equations

Derivatives?

3. The attempt at a solution

Okay so I directly have a first question regarding the problem: can I treat the set as a function or is that nonsense? I think the derivatives of that expression should tell me where the slope of the function is zero and therefore potentially reaching an extremum (or not), right?

So here we go:

(d/dx)(x2 + x + 1)/(x2 + 1) = [(2x + 1)(x2 + 1) - (x2 + x + 1)(2x)]/(x2 + 1)2 = (-x2 + 1)/(x2 + 1)2

which is equal to 0 for x = {-1;1}. Now I take the second derivative of the set to check if it is also equal to 0 for x = {-1;1}, and if it is not then those values of x would mark two extrema.

(d/dx)(-x2 + 1)/(x2 + 1)2 = [2x(x2 - 3)]/(x2 + 1)3

which is equal to 1/2 for x = -1 and -1/2 for x = 1, which would that the set has a minimum at x = 1 and a maximum at x = -1. But I know from the graph that that is wrong, although the values for x are right :/

Where is my mistake? Is that to begin with the right way to proceed? And if yes, are there also other methods? Do you have any remark about such problems, especially about the difference between extrema of sets and of functions?

Julien.

Last edited: Apr 7, 2016
2. Apr 7, 2016

### JulienB

Oh I think I just wrote the method wrong. If f'(x0 ) = 0 and f''(x0) < 0, then there is a maximum at x = x0? That would make sense with the problem above.

Julien.

Last edited by a moderator: Apr 8, 2016
3. Apr 7, 2016

### Ray Vickson

Your conclusions are exactly backwards: $x = -1$ is a (global) minimum, while $x = 1$ is a global max.

4. Apr 8, 2016