2021-M2-05

設 $y=mx+c$ 為 $y=r(x)$ 的斜漸近線。

$\begin{array}{rcl}
m & = & \dlim_{x \to \infty} \left(\dfrac{r(x)}{x} \right) \\
m & = & \dlim_{x \to \infty} \dfrac{x^3-x^2-2x+3}{x(x-1)^2} \\
m & = & \dlim_{x \to \infty} \dfrac{x^3-x^2-2x+3}{x^3-2x^2+x} \\
m & = & \dlim_{x \to \infty} \dfrac{(x^3-x^2-2x+3)/x^3}{(x^3-2x^2+x)/x^3} \\
m & = & \dlim_{x \to \infty} \dfrac{1-\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^2}} \\
m & = & 1
\end{array}$

另外，

$\begin{array}{rcl}
c & = & \dlim_{x \to \infty} [r(x)-mx] \\
c & = & \dlim_{x\to\infty} \left(\dfrac{x^3-x^2-2x+3}{(x-1)^2}-x\right) \\
c & = & \dlim_{x\to\infty} \dfrac{x^3-x^2-2x+3-x(x-1)^2}{x^2-2x+1} \\
c & = & \dlim_{x\to\infty} \dfrac{x^3-x^2-2x+3-x^3+2x^2-x}{x^2-2x+1} \\
c & = & \dlim_{x\to\infty} \dfrac{x^2-3x+3}{x^2-2x+1} \\
c & = & \dlim_{x\to\infty} \dfrac{(x^2-3x+3)/x^2}{(x^2-2x+1)/x^2} \\
c & = & \dlim_{x\to\infty} \dfrac{1-\frac{3}{x}+\frac{3}{x^2}}{1-\frac{2}{x}+\frac{1}{x^2}} \\
c & = & 1
\end{array}$

$\therefore y=x+1$ 為一斜漸近線。

$\begin{array}{rcl}
\dfrac{d}{dx}r(x) & = & \dfrac{x^3-3x^2+4x-4}{(x-1)^3} \\
\dfrac{d^2}{dx^2}r(x) & = & \dfrac{(x-1)^3(3x^2-6x+4)-(x^3-3x^2+4x-4)\times3(x-1)^2}{(x-1)^6} \\
\dfrac{d^2}{dx^2}r(x) & = & \dfrac{(x-1)^2[(x-1)(3x^2-6x+4)-3(x^3-3x^2+4x-4)]}{(x-1)^6} \\
\dfrac{d^2}{dx^2}r(x) & = & \dfrac{3x^3-6x^2+4x-3x^2+6x-4-3x^3+9x^2-12x+12}{(x-1)^4} \\
\dfrac{d^2}{dx^2}r(x) & = & \dfrac{-2x+8}{(x-1)^4}
\end{array}$

對於拐點，

$\begin{array}{rcl}
\dfrac{d^2}{dx^2}r(x) & = & 0 \\
\dfrac{-2x+8}{(x-1)^4} & = & 0 \\
-2x+8 & = & 0 \\
x & = 4
\end{array}$

$\begin{array}{|l|c|c|c|c|c|} \hline
x & x < 1 & x = 1 & 1 < x < 4 & x=4 & x > 4 \\ \hline
\dfrac{d^2}{dx^2}r(x) & +ve & \text{不被定義} & +ve & 0 & -ve \\ \hline
r(x) & \text{向上凹} & \text{不被定義} & \text{向上凹} & \text{拐點} & \text{向下凹} \\ \hline
\end{array}$

由此，$y=r(x)$ 只有一個拐點。

所以，我同意該宣稱。