Finding global maximum and minimum of a given function

Question

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?

Answer 1

we get $\frac{1}{3}\le f(x)\le 3$ since $3-f(x)=\frac{2(x+2)^2}{x^2+2x+4}$ and $f(x)-\frac{1}{3}=\frac{2}{3}\frac{(x-2)^2}{x^2+2x+4}$

Answer 2

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

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 $f(t)=1-\frac {4}{t+2}$.