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O último item enumerado não está alinhado

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O último item enumerado não está alinhado

alguém sabe por que o último item (e os itens adicionados após o dia 7) não está alinhado com os itens anteriores? Eu não entendo...

\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}

Todos os outros itens se alinham como o 6º, exceto o 7º item.

Responder1

Existem erros no seu documento, então na verdade ele não compila. É por isso que na saída (na qual você não deve confiar porque há erros) o último item da enumeração não está alinhado corretamente.

  • Na linha 122, você escreveu cm^3em modo texto, mas o expoente escrito com ^3deve estar em modo matemático. Para compor unidades, a melhor maneira é usarsiunitx: você pode simplesmente substituir cm^3por \unit{cm^3}.
  • Na linha 137, existem polinômios com expoentes em modo texto. Isso deve estar no modo matemático.
  • Na linha 147, você escreveu m^3em modo texto, então ocorre o mesmo erro que com cm^3.

Aqui está uma versão corrigida do seu exemplo.

\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}

Agora ele compila e o último item da enumeração está alinhado corretamente na saída.

Acho que devo salientar também que ainda existem erros tipográficos no seu documento, mesmo que não resultem em erros de LaTeX. Aqui estão duas sugestões.

  • xàs vezes é escrito em modo texto quando pretende ser uma variável matemática, por exemplo, na primeira linha do item 7 na enumeração. Quando “x” denota uma quantidade matemática, deve sempre ser escrito em modo matemático.
  • Muitas das linhas com manipulações algébricas e outros conteúdos matemáticos seriam muito mais fáceis de ler se você usasse ambientes matemáticos de exibição (por exemplo equation, alignou gather) para escrevê-las.

Responder2

O problema é causado por m ^ 3 no sétimo item. Isso é um erro; digitar m^3para obter m ao cubo só funciona no modo matemático. Use m\textsuperscript{3}em vez disso.

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