Let's represent the following function (1 unit $=2 \mathrm{~cm}$):

$$
y=\frac{x^{2}-x-1}{x^{2}+x+1}
$$
# I. Solution:

a) Since the denominator

$$
x^{2}+x+1=\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4}>0
$$

for all real $x$, the function is defined over the entire real line.

b) Let's find the zeros of the curve, i.e., the points where the curve intersects the $x$-axis. $y=0$ if the numerator

$$
x^{2}-x-1=0
$$

or if

$$
x_{1}=\frac{1-\sqrt{5}}{2} \quad \text { and } \quad x_{2}=\frac{1+\sqrt{5}}{2} \text {. }
$$

(The denominator, as we have seen, is never equal to 0.)
c) When $x=0$, $y=-1$. If $x \neq 0$, our function can be written in the form

$$
y=1-2 \frac{x+1}{x^{2}+x+1}=1-2 \frac{1+\frac{1}{x}}{x+1+\frac{1}{x}}
$$

If here the absolute value of $x$ is sufficiently large (and certainly greater than 1), then the absolute value of the numerator will certainly be less than, for example, 2, while the absolute value of the denominator will become arbitrarily large. The absolute value of the fraction will therefore become arbitrarily small. Thus, if the absolute value of $x$ exceeds all bounds, then $y$ will approach the value 1 with arbitrary precision. Otherwise, the fraction $\frac{x+1}{x^{2}+x+1}$ is positive if $x>-1$, and negative if $x<-1$, so (also considering the factor of -2) for negative $x$ the curve of the function approaches the line $y=1$ from above, and for positive $x$ from below (asymptote).

d) Let's find the extreme values of the function. For this purpose, express $x$ in terms of $y$. Since the denominator is never 0, we have

$$
y\left(x^{2}+x+1\right)=x^{2} y+x y+y=x^{2}-x-1
$$

or

$$
x^{2}(y-1)+x(y+1)+(y+1)=0
$$

from which

$$
x=\frac{-(y+1) \pm \sqrt{(y+1)^{2}-4\left(y^{2}-1\right)}}{2(y-1)}
$$

$x$ is only real if the discriminant

$$
(y+1)^{2}-4\left(y^{2}-1\right)=-3 y^{2}+2 y+5 \geq 0
$$

Since the roots of the equation

$$
-3 y^{2}+2 y+5=0
$$

are

$$
y_{1}=-1, \quad \text { and } \quad y_{2}=\frac{5}{3}
$$

the inequality (1) holds if

$$
-1 \leq y \leq \frac{5}{3}
$$

Thus, the extreme values of $y$ are:

$$
\begin{array}{lll}
y_{\min }=-1 & \text { when } & x=0 \\
y_{\max }=\frac{5}{3}, & \text { when } & x=-2
\end{array}
$$

By substitution, we obtain the following value table

| $x$ | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | $\frac{11}{7}$ | $\frac{5}{3}$ | 1 | -1 | $-\frac{1}{3}$ | $\frac{1}{7}$ | $\frac{5}{13}$ |

![](https://cdn.mathpix.com/cropped/2024_05_02_8272afb498e6904a7c4cg-2.jpg?height=583&width=729&top_left_y=175&top_left_x=687)

Károly Bauer (Pécs, Mining Technology IV. o. t.)

## II. Solution:

The extreme values can also be determined as follows. Transform the function:

$$
\begin{aligned}
y=\frac{x^{2}-x-1}{x^{2}+x+1} & =1-2 \frac{x+1}{x(x+1)+1}=1-\frac{2}{x+\frac{2}{x+1}}= \\
& =1-\frac{2}{(x+1)+\frac{1}{x+1}-1}
\end{aligned}
$$

$y$ is minimal when the second term is maximal, or when the denominator of the second term is minimal in the set of non-negative numbers. But $x+1+\frac{1}{x+1}$ is minimal in the set of non-negative numbers when $x+1=\frac{1}{x+1}=1$, i.e., $x=0$, in which case $y_{\min }=-1$.

$y$ is maximal when the second term is minimal, or when the denominator is maximal in the set of negative numbers. But $x+1+\frac{1}{x+1}$ is maximal in the set of negative numbers when $x+1=\frac{1}{x+1}=-1$, from which $x=-2$ and $y_{\max }=\frac{5}{3}$.

Imre Csiszár (Bp. I., Petőfi g. II. o. t.)

## III. Solution:

A third method for determining the extreme value is as follows:

$$
y=\frac{2 x^{2}}{x^{2}+x+1}-1=\frac{2}{1+\frac{1}{x}+\frac{1}{x^{2}}}-1=\frac{2}{\left(\frac{1}{x}+\frac{1}{2}\right)^{2}+\frac{3}{4}}-1
$$

$y$ is maximal when the first term is maximal, or when the denominator of the first term is minimal. $\left(\frac{1}{x}+\frac{1}{2}\right)^{2}$ is a non-negative quantity, which is minimal when its value is 0, i.e., when $x=-2$, in which case $y_{\max }=\frac{5}{3}$.

$y$ is minimal when the denominator of the first term is maximal, or when $\frac{1}{x}$ is maximal. The latter will exceed all bounds as $x$ approaches 0; thus, the value of the fraction (which is always positive) will also approach 0, so $y_{\min }=-1$.