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最後一個枚舉項未對齊

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最後一個枚舉項未對齊

有誰知道為什麼最後一個項目（以及第七個之後添加的項目）與之前的項目不一致？我不明白...

\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{float}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[margin=0.5in]{geometry}

\title{Unit 1 Assessment, Part 2}
\date{May 2022}

\begin{document}

\maketitle

\begin{enumerate}
  \item x - intercepts of the quadratic function $f(x)=-x^{2}+7x-6$ is at: \\
  $0 = -x^{2} + 7x - 6 = (x - 1)(-x + 6) \\
  x = 1,x = 6$ \\
  The vertical asymptotes of the reciprocal function $g(x)=\frac{1}{-x^{2}+7x-6}$: \\
  $x = 1,x = 6$ \\
  \\
  The horizontal asymptote of the reciprocal is y = 0  since all reciprocal functions have a horizontal asymptote at y = 0. \\
  \\
  The interval of increase of the quadratic function is $(-\infty,3.5)$, and the interval of decrease of the quadratic function is $(3.5,\infty)$. Therefore, the interval of decrease of the reciprocal function is $(-\infty,1)\cup(1,3.5)$, and the interval of increase of the reciprocal function is $(3.5,6)\cup(6,\infty)$.\\
  \\
  The quadratic function has a maximum point at x = 3.5, therefore the reciprocal has a minimum point at x = 3.5. \\
  \\
  The positive interval of the quadratic function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$. Therefore, the positive interval of the reciprocal function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$.\\
  \begin{figure}[H]
    \centering
    \includegraphics[scale=0.5]{unit1part2a.png}
    \caption{Graph: $f(x)=-x^{2}+7x-6$ \& $g(x)=\frac{1}{-x^{2}+7x-6}$.}
  \end{figure}

  \item \begin{enumerate}
    \item $f(x) = \frac{-2x - 5}{3x + 18}$
    \begin{center}
      \begin{tabular}{|c|c|c|c|c|}
          \hline
          Vertical asymptotes & Horizontal asymptotes & x - intercept & y - intercept & Domain \\
          \hline
          x = -6 & $y =  -\frac{2}{3}$ & $(-\frac{5}{2},0)$ & $(0,-\frac{5}{18})$ & $D = \{x\in\mathbb{R}|x\neq-6\}$ \\
          \hline
      \end{tabular}
    \end{center}
    \begin{figure}[H]
      \centering
      \includegraphics[scale=0.5]{unit1part2b.png}
      \caption{Graph: $f(x) = \frac{-2x - 5}{3x + 18}$.}
    \end{figure}

    \item Positive interval: $(-6,-2.5)$ \\
     Negative intervals: $(-\infty,-6)\cup(-2.5,\infty)$
  \end{enumerate}

  \item Find the real roots of the following rational equations.
  \begin{enumerate}
    \item $\frac{-7x}{9x + 11} - 12 = \frac{1}{x} \\
    \frac{-7x}{9x + 11} = \frac{1 + 12x}{x} \\
    (-7x)(x) = (9x + 11)(1 + 12x) \\
    -7x^2 = 9x + 108x^2 + 11 + 132x \\
    115x^2 + 141x + 11 = 0 \\
    x = \frac{-141\pm\sqrt{141^2 - (4)(115)(11)}}{(2)(115)} \\
    x \approx -0.08,x \approx -1.14$

    \item $\frac{x - 1}{x + 2} = \frac{3x + 8}{5x - 1} \\
    (x - 1)(5x - 1) = (x + 2)(3x + 8) \\
    5x^2 - x - 5x + 1 = 3x^2 + 8x + 6x + 16 \\
    2x^2 - 20x - 15 = 0 \\
    x = \frac{-(-20)\pm\sqrt{(-20)^2 - (4)(2)(-15)}}{(2)(2)} \\
    x = \frac{10\pm\sqrt{130}}{2}$
  \end{enumerate}

  \item $8x - 3 \leq 2x+1 \leq 17x - 8 \\
  8x - 3 \leq 2x+1 \\
  6x \leq 4 \\
  x \leq \frac{2}{3} \\
  2x+1 \leq 17x - 8 \\
  9 \leq 15x \\
  x \geq \frac{3}{5} \\
  \frac{3}{5} \leq x \leq \frac{2}{3}$

  \item $\frac{5x + 4}{x - 11} < \frac{5x - 7}{x + 13} \\
  \frac{5x + 4}{x - 11} - \frac{5x - 7}{x + 13} < 0 \\
  \frac{(5x + 4)(x + 13) - (5x - 7)(x - 11)}{(x - 11)(x + 13)} < 0 \\
  \frac{5x^2 + 65x + 4x + 52 - 5x^2 + 55x + 7x - 77}{(x - 11)(x + 13)} < 0 \\
  \frac{121x - 25}{(x - 11)(x + 13)} < 0 \\
  \text{Critical numbers:} \\
  121x - 25 = 0 \\
  x = \frac{25}{121} \\
  x - 11 = 0 \\
  x = 11 \\
  x + 13 = 0 \\
  x = -13$ \\
  \begin{tabular}{|c|c|c|c|c|c|}
    \hline
    Intervals & Test value x & 121x - 25 & x - 11 & x + 13 & $\frac{121x - 25}{(x - 11)(x + 13)}$ \\
    \hline
    $x < -13$ & -14 & - & - & - & - \\
    \hline
    $-13 < x < \frac{25}{121}$ & 0 & - & - & + & + \\
    \hline
    $ \frac{25}{121} < x < 11$ & 10 & + & - & + & - \\
    \hline
    $x > 11$ & 12 & + & + & + & + \\
    \hline
  \end{tabular} \\
  \\
  Therefore, the solution is $(-\infty,-13)\cup(\frac{25}{121},11)$.

  \item $(3 + x)(5 + x)(7 + x) = 693 \\
  (15 + 3x + 5x + x^2)(7 + x) = 693 \\
  105 + 15x + 56x + 8x^2 + 7x^2 + x^3 - 693 = 0 \\
  x^3 + 15x^2 + 71x - 588 = 0 \\
  \because 4^3 + 15(4)^2 + 71(4) - 588 = 0 \\
  \therefore x - 4$ is a factor. \\
  Devide $x^3 + 15x^2 + 71x - 588$ by x - 4: \\
  $(x - 4)(x^2 + 19x + 127) = 0 \\
  \because 19^2 - (4)(1)(127) < 0 \\
  \therefore x^2 + 19x + 127 = 0$ has no real solution. \\
  When x - 4 = 0, x = 4 \\
  The value of x is 4 will produce a box with a volume of 693 cm^3.

  \item Let x represent the width in meters. \\
  $(3x + 1)(2x - 5)x \geq 8436 \\
  6x^3 - 13x^2 - 5x - 8436 \geq 0 \\
  \because 6(12)^3 - 13(12)^2 - 5(12) - 8436 = 0 \\
  \therefore x - 12$ is a factor. \\
  Devide $6x^3 - 13x^2 - 5x - 8436$ by x - 12: \\
  $(x - 12)(6x^2 + 59x + 703) \geq 0 $\\
  Critical number: \\
  $\because 59^2 - (4)(6)(703) < 0 \\
  \therefore 6x^2 + 59x + 703 = 0$ has no real solution. \\
  When x - 12 = 0, x = 12 \\
  \begin{tabular}{|c|c|c|c|c|}
    \hline
    Intervals & Test value x & x - 12 & 6x^2 + 59x + 703 & (x - 12)(6x^2 + 59x + 703) \\
    \hline
    $x < 12$ & 11 & - & + & - \\
    \hline
    $x > 12$ & 13 & + & + & + \\
    \hline
  \end{tabular} \\
  $D = \{x\in\mathbb{R}|x\geq12\} \\
  3(12) + 1  = 37 \\
  2(12) - 5 = 19 \\$
  When the length is greater or equal to 37 m, the height is greater or equal to 19 m, and the width is greater or equal to 12 m, the volume of the container is at least 8436 m^3.
\end{enumerate}

\end{document}

除第七項外，所有其他項均與第六項一樣對齊。

答案1

您的文檔中有錯誤，所以實際上它無法編譯。這就是為什麼在輸出中（無論如何您都不應該相信它，因為存在錯誤）枚舉的最後一項未正確對齊。

  • 在第 122 行，您cm^3以文字模式編寫，但用 編寫的指數^3必須是數學模式。要排版單位，最好的方法是使用siunitx: 你可以簡單地替換cm^3為\unit{cm^3}.
  • 第 137 行，有文字模式下帶指數的多項式。那應該是數學模式。
  • 在第 147 行，您m^3以文字模式編寫，因此會發生與cm^3.

這是您的範例的更正版本。

\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{float}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[margin=0.5in]{geometry}
\usepackage{siunitx}

\title{Unit 1 Assessment, Part 2}
\date{May 2022}

\begin{document}

\maketitle

\begin{enumerate}
  \item x - intercepts of the quadratic function $f(x)=-x^{2}+7x-6$ is at: \\
  $0 = -x^{2} + 7x - 6 = (x - 1)(-x + 6) \\
  x = 1,x = 6$ \\
  The vertical asymptotes of the reciprocal function $g(x)=\frac{1}{-x^{2}+7x-6}$: \\
  $x = 1,x = 6$ \\
  \\
  The horizontal asymptote of the reciprocal is y = 0  since all reciprocal functions have a horizontal asymptote at y = 0. \\
  \\
  The interval of increase of the quadratic function is $(-\infty,3.5)$, and the interval of decrease of the quadratic function is $(3.5,\infty)$. Therefore, the interval of decrease of the reciprocal function is $(-\infty,1)\cup(1,3.5)$, and the interval of increase of the reciprocal function is $(3.5,6)\cup(6,\infty)$.\\
  \\
  The quadratic function has a maximum point at x = 3.5, therefore the reciprocal has a minimum point at x = 3.5. \\
  \\
  The positive interval of the quadratic function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$. Therefore, the positive interval of the reciprocal function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$.\\
  \begin{figure}[H]
    \centering
    \includegraphics[scale=0.5]{example-image-a}
    \caption{Graph: $f(x)=-x^{2}+7x-6$ \& $g(x)=\frac{1}{-x^{2}+7x-6}$.}
  \end{figure}

  \item \begin{enumerate}
    \item $f(x) = \frac{-2x - 5}{3x + 18}$
    \begin{center}
      \begin{tabular}{|c|c|c|c|c|}
          \hline
          Vertical asymptotes & Horizontal asymptotes & x - intercept & y - intercept & Domain \\
          \hline
          x = -6 & $y =  -\frac{2}{3}$ & $(-\frac{5}{2},0)$ & $(0,-\frac{5}{18})$ & $D = \{x\in\mathbb{R}|x\neq-6\}$ \\
          \hline
      \end{tabular}
    \end{center}
    \begin{figure}[H]
      \centering
      \includegraphics[scale=0.5]{example-image-b}
      \caption{Graph: $f(x) = \frac{-2x - 5}{3x + 18}$.}
    \end{figure}

    \item Positive interval: $(-6,-2.5)$ \\
     Negative intervals: $(-\infty,-6)\cup(-2.5,\infty)$
  \end{enumerate}

  \item Find the real roots of the following rational equations.
  \begin{enumerate}
    \item $\frac{-7x}{9x + 11} - 12 = \frac{1}{x} \\
    \frac{-7x}{9x + 11} = \frac{1 + 12x}{x} \\
    (-7x)(x) = (9x + 11)(1 + 12x) \\
    -7x^2 = 9x + 108x^2 + 11 + 132x \\
    115x^2 + 141x + 11 = 0 \\
    x = \frac{-141\pm\sqrt{141^2 - (4)(115)(11)}}{(2)(115)} \\
    x \approx -0.08,x \approx -1.14$

    \item $\frac{x - 1}{x + 2} = \frac{3x + 8}{5x - 1} \\
    (x - 1)(5x - 1) = (x + 2)(3x + 8) \\
    5x^2 - x - 5x + 1 = 3x^2 + 8x + 6x + 16 \\
    2x^2 - 20x - 15 = 0 \\
    x = \frac{-(-20)\pm\sqrt{(-20)^2 - (4)(2)(-15)}}{(2)(2)} \\
    x = \frac{10\pm\sqrt{130}}{2}$
  \end{enumerate}

  \item $8x - 3 \leq 2x+1 \leq 17x - 8 \\
  8x - 3 \leq 2x+1 \\
  6x \leq 4 \\
  x \leq \frac{2}{3} \\
  2x+1 \leq 17x - 8 \\
  9 \leq 15x \\
  x \geq \frac{3}{5} \\
  \frac{3}{5} \leq x \leq \frac{2}{3}$

  \item $\frac{5x + 4}{x - 11} < \frac{5x - 7}{x + 13} \\
  \frac{5x + 4}{x - 11} - \frac{5x - 7}{x + 13} < 0 \\
  \frac{(5x + 4)(x + 13) - (5x - 7)(x - 11)}{(x - 11)(x + 13)} < 0 \\
  \frac{5x^2 + 65x + 4x + 52 - 5x^2 + 55x + 7x - 77}{(x - 11)(x + 13)} < 0 \\
  \frac{121x - 25}{(x - 11)(x + 13)} < 0 \\
  \text{Critical numbers:} \\
  121x - 25 = 0 \\
  x = \frac{25}{121} \\
  x - 11 = 0 \\
  x = 11 \\
  x + 13 = 0 \\
  x = -13$ \\
  \begin{tabular}{|c|c|c|c|c|c|}
    \hline
    Intervals & Test value x & 121x - 25 & x - 11 & x + 13 & $\frac{121x - 25}{(x - 11)(x + 13)}$ \\
    \hline
    $x < -13$ & -14 & - & - & - & - \\
    \hline
    $-13 < x < \frac{25}{121}$ & 0 & - & - & + & + \\
    \hline
    $ \frac{25}{121} < x < 11$ & 10 & + & - & + & - \\
    \hline
    $x > 11$ & 12 & + & + & + & + \\
    \hline
  \end{tabular} \\
  \\
  Therefore, the solution is $(-\infty,-13)\cup(\frac{25}{121},11)$.

  \item $(3 + x)(5 + x)(7 + x) = 693 \\
  (15 + 3x + 5x + x^2)(7 + x) = 693 \\
  105 + 15x + 56x + 8x^2 + 7x^2 + x^3 - 693 = 0 \\
  x^3 + 15x^2 + 71x - 588 = 0 \\
  \because 4^3 + 15(4)^2 + 71(4) - 588 = 0 \\
  \therefore x - 4$ is a factor. \\
  Devide $x^3 + 15x^2 + 71x - 588$ by x - 4: \\
  $(x - 4)(x^2 + 19x + 127) = 0 \\
  \because 19^2 - (4)(1)(127) < 0 \\
  \therefore x^2 + 19x + 127 = 0$ has no real solution. \\
  When x - 4 = 0, x = 4 \\
  The value of x is 4 will produce a box with a volume of 693 \unit{cm^3}.

  \item Let x represent the width in meters. \\
  $(3x + 1)(2x - 5)x \geq 8436 \\
  6x^3 - 13x^2 - 5x - 8436 \geq 0 \\
  \because 6(12)^3 - 13(12)^2 - 5(12) - 8436 = 0 \\
  \therefore x - 12$ is a factor. \\
  Devide $6x^3 - 13x^2 - 5x - 8436$ by x - 12: \\
  $(x - 12)(6x^2 + 59x + 703) \geq 0 $\\
  Critical number: \\
  $\because 59^2 - (4)(6)(703) < 0 \\
  \therefore 6x^2 + 59x + 703 = 0$ has no real solution. \\
  When x - 12 = 0, x = 12 \\
  \begin{tabular}{|c|c|c|c|c|}
    \hline
    Intervals & Test value x & x - 12 & $6x^2 + 59x + 703$ & $(x - 12)(6x^2 + 59x + 703)$ \\
    \hline
    $x < 12$ & 11 & - & + & - \\
    \hline
    $x > 12$ & 13 & + & + & + \\
    \hline
  \end{tabular} \\
  $D = \{x\in\mathbb{R}|x\geq12\} \\
  3(12) + 1  = 37 \\
  2(12) - 5 = 19 \\$
  When the length is greater or equal to 37 m, the height is greater or equal to 19 m, and the width is greater or equal to 12 m, the volume of the container is at least 8436 \unit{m^3}.
\end{enumerate}

\end{document}

現在，它確實可以編譯，並且枚舉的最後一項在輸出中正確對齊。

我想我還應該指出，您的文件中仍然存在印刷錯誤，即使它們不會導致 LaTeX 錯誤。這裡有兩個建議。

  • x當它是一個數學變數時，有時會以文字模式編寫，例如在枚舉中第 7 項的第一行。當“x”表示數學量時，應始終以數學模式書寫。
  • equation如果您使用顯示數學環境（例如、align或gather）來編寫包含代數運算和其他數學內容的許多行，那麼它們將更容易閱讀。

答案2

問題是由第七項中的 m^3 引起的。這是一個錯誤；鍵入m^3以獲得 m 的立方僅適用於數學模式。代替使用m\textsuperscript{3}。

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