$13 \cdot 34$ For $-4<x<1$, the expression $\frac{x^{2}-2 x+2}{2 x-2}$
(A) has neither a maximum nor a minimum value.
(B) has a minimum value of 1.
(C) has a maximum value of 1.
(D) has a minimum value of -1.
(E) has a maximum value of -1.
(9th American High School Mathematics Examination, 1958)
[Solution] The expression is transformed to $\frac{x^{2}-2 x+2}{2 x-2}=\frac{1}{2}\left(x-1+\frac{1}{x-1}\right)$.
Given $-4<x<1$, we have $-5<x-1<0$.
Let $t=x-1$, then $t+\frac{1}{t}$.
When $t=\frac{1}{t}$, i.e., $t^{2}=1$, there is a maximum value.
Thus, $-5<t<0$, so when $t=-1$, the original expression has a maximum value of -1.
Therefore, the answer is $(E)$.
Note: A similar problem is
For $x \geqslant 0$, the minimum value of the algebraic expression $\frac{4 x^{2}+8 x+13}{6(1+x)}$ is
(A) 1.
(B) 2.
(C) $\frac{25}{12}$.
(D) $\frac{13}{6}$.
(E) $\frac{34}{5}$.
(16th American High School Mathematics Examination, 1965)
[Solution] Let $y=\frac{4 x^{2}+8 x+13}{6(1+x)}=\frac{4(x+1)^{2}+9}{6(1+x)}, x \geqslant 0$.
Set $x+1=t$, then $y=\frac{4 t^{2}+9}{6 t}, \quad t \geqslant 1$.
That is,
$$
4 t^{2}-6 t y+9=0 .
$$

Since $t$ is a real number, $\Delta=36 y^{2}-4 \cdot 4 \cdot 9 \geqslant 0$, i.e., $y^{2}-4 \geqslant 0$. Thus, $y \geqslant 2$ or $y \leqslant-2$ (for $x \geqslant 0$, $y \geqslant 0$, so the latter is not applicable). From $y \geqslant 2$, we know the minimum value of $y$ is 2, which gives $t=\frac{3}{2}$, or $x=\frac{1}{2}$. Therefore, the answer is (B).