They asked for the maximum and minimum of the function

$$
y=\frac{x^{2}-2 x+1}{2 x^{2}+2 x+1}
$$
Find the maximum and minimum of the functions

$$
y=\frac{x^{2}+1}{x^{2}+4 x+3} \quad \text { and } \quad y=\frac{x^{2}-2 x+1}{2 x^{2}+2 x+1}
$$

# First Solution. 

Write the first equation in the form of decreasing powers of $x$:

$$
(1-y) x^{2}-4 y x+(1-3 y)=0
$$

The roots of this equation are real if

$$
y^{2}-4 y-1 \geqq 0
$$

or if

$$
[y-(2+\sqrt{5})][y-(2-\sqrt{5})] \geqq 0
$$

This follows if

$$
y \geqq(2+\sqrt{5}) \geqq(2-\sqrt{5})
$$

or if

$$
y \leqq(2-\sqrt{5}) \leqq(2+\sqrt{5})
$$

In the sequence of permissible values of $y$, $y^{\prime}=2-\sqrt{5}$ is the upper bound, i.e., the maximum, and $y^{\prime \prime}=2+\sqrt{5}$ is the lower bound, i.e., the minimum.

## Second Solution.

Let $x_{1}$ and $x_{2}$ be two values of $x$ at which $y$ takes the same value; for the second function, we have

$$
\begin{gathered}
\frac{x_{1}^{2}-2 x_{1}+1}{2 x_{1}^{2}+2 x_{1}+1}=\frac{x_{2}^{2}-2 x_{2}-1}{2 x_{2}^{2}+2 x_{2}+1} \\
2 x_{1}^{2} x_{2}^{2}+2 x_{1}^{2} x_{2}+x_{1}^{2}-4 x_{1} x_{2}^{2}-4 x_{1} x_{2}-2 x_{1}+2 x_{2}^{2}+2 x_{2}+1= \\
=2 x_{2}^{2} x_{1}^{2}+2 x_{2}^{2} x_{1}+x_{2}^{2}-4 x_{2} x_{1}^{2}-4 x_{2} x_{1}-2 x_{2}+2 x_{1}^{2}+2 x_{1}+1 \\
2 x_{1} x_{2}\left(x_{1}-x_{2}\right)+\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)-2\left(x_{1}-x_{2}\right)+4 x_{1} x_{2}\left(x_{1}-x_{2}\right)- \\
-2\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)-2\left(x_{1}-x_{2}\right)=0 \\
6 x_{1} x_{2}-\left(x_{1}+x_{2}\right)-4=0
\end{gathered}
$$

If now

$$
x_{1}=x_{2}=x
$$

then

$$
6 x^{2}-2 x-4=0
$$

from which

$$
x=\left\{\begin{array}{l}
=1 \\
=-\frac{2}{3}
\end{array}\right.
$$

The corresponding values of $y$ are

$$
\begin{aligned}
& y_{\min }=0 \\
& y_{\max }=5
\end{aligned}
$$

(Aladár Visnya, fr. VII. o. t. Pécs.)