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Find the global maximum and the global minimum of the function $f$ on $\mathbb R$, where $f(x)= \frac {x^2-2x+4}{x^2+2x+4}, x \in \mathbb R$

My approach: I used the higher order derivative test, i.e. I calculated $f'(x), f''(x)$, found the points where $f'(x)=0$, and then showed that $f''(.)$ is greater or less than $0$ at those points. In this way, I found the LOCAL maximum and minimum.

Now, I know a result that if $f:I \rightarrow \mathbb R$ has a local maximum or minimum at a point $c \in I$ then $c$ is a global maximum on $N(c, \delta) \cap I$, where $\delta >0$

Is there any way to use this result to show that the local extremum I found here are global extremum on $\mathbb R$?

The method used in my book to solve this sum uses supremum and infimum, and I can't understand that working because the results are not given in the book. Could someone please give me references (pdfs or links) for global maximum or minimum involving supremum and infimum?

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    $\begingroup$ You can't, in general, hope to locate global extrema by finding critical points. You may have plenty of local extrema but no global extrema at all. You need, in general, to consider the whole function, i.e. you should perform a careful study of the function in its domain of definition. $\endgroup$ – Siminore Oct 31 '14 at 10:19
  • $\begingroup$ How do I do that? Using supremum and infimum? Can you please suggest a link to study that from? $\endgroup$ – Diya Oct 31 '14 at 10:22
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    $\begingroup$ As a rule, you need to study the behavior of the function: monotonicity, limits at infinity, and so on... $\endgroup$ – Siminore Oct 31 '14 at 12:30
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we get $\frac{1}{3}\le f(x)\le 3$ since $3-f(x)=\frac{2(x+2)^2}{x^2+2x+4}$ and $f(x)-\frac{1}{3}=\frac{2}{3}\frac{(x-2)^2}{x^2+2x+4}$

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$$f(x)= \frac {x^2-2x+4}{x^2+2x+4}=1-\frac {4x}{x^2+2x+4}$$

Hence $f(0)=1$ and if $x\neq 0$, $$f(x)=1-\frac {4}{x+\frac{4}{x}+2}$$

Let $t=x+\frac{4}{x}\in(-\infty,-4]\cup[4,\infty)$

you only need to consider $f(t)=1-\frac {4}{t+2}$.

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  • $\begingroup$ I have already done that. I used the higher order derivative test directly to find the LOCAL maximum and minimum. My question is, is there any way to prove it is the GLOBAL maximum and minimum? $\endgroup$ – Diya Oct 31 '14 at 10:13

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