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我是 LaTeX 新手。这是我的第一个脚本,一个练习。我尝试将SCfigure
已经绘制的 4 个数字 () 放在KmPlot
其对应的方程旁边,但它们总是出现在下一页。有什么建议吗?
\documentclass[a4paper]{report}
\usepackage[spanish,activeacute]{babel}
\usepackage{amssymb,amsmath,amsbsy,amstext,amscd,amsxtra,amsopn}
\usepackage{fullpage}
\usepackage{graphicx,wrapfig,sidecap}
\title{Matem\'atica - Parcial 1}
\author{Ignacio Guerrero\\
Leg.Nº 57.866\\
Univ. Blas Pascal}
\begin{document}
\maketitle
\subsection*{1.a}
\begin{equation*}\label{xx}
\begin{split}
2(x-1)-3(x-4)&=4x\\
2x-2-3x+12&=4x\\
2x-3x-4x&=-10\\
\frac{-5x}{-5}& =\frac{-10}{-5}\\
\\
\boxed{x=-2}
\end{split}
\end{equation*}
\subsection*{1.b}
\begin{equation*}\label{xx}
\begin{split}
x^2-9&=0\\
x^2&=9\\
x&=\pm \sqrt{9}\\
\\
\boxed{x_1=3} & \qquad \boxed{ x_2=-3}
\end{split}
\end{equation*}
\subsection*{2}
\begin{equation*}\label{xx}
\begin{aligned}
(x_1,y_1)=(-2,0) & \qquad (x_2,y_2)=(1,3)
\end{aligned}
\end{equation*}
\begin{equation*}\label{xx}
\begin{split}
a&=\frac{y_1-y_2}{x_1-x_2}\\
\\
a_1&=\frac{0-3}{-2-1}=\frac{-3}{-3}=1\\
a_2&=\frac{2-k}{-1+6}=\frac{2-k}{5}\\
\\
a_1&=a_2\\
1&=\frac{2-k}{5}\\
5&=2-k\\
\\
\boxed
{k=-3}
\end{split}
\end{equation*}
\subsection*{3.a}
\begin{SCfigure}
\includegraphics[width=7cm,height=7cm]{exer3a}
\caption{La gr\'afica nos muestra que ambas rectas son paralelas dado que compaten la misma pendiente $(a=-3x)$ pero cortan al eje $x$ en distintos puntos. Por lo tanto, el sistema es incompatible.}
\label{3.a}
\end{SCfigure}
\begin{equation*}\label{xx}
\begin{aligned}
\begin{cases}
3x+y&=3\\
x+\frac{1}{3}y&=-1
\end{cases}
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{split}
y_1&=-3z+3\\
y_2&=3(-x-1)=-3x-3
\end{split}
\end{equation*}
\subsection*{3.b}
\begin{equation*}
\begin{split}
\begin{cases}
2x-5y&=4\\
x-3y&=2
\end{cases}
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
y_1&=\frac{-2x+4}{-5}\\
y_2&=\frac{-x+2}{-3}
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
x-3y&=2\\
x&=3y+2\\
\\
2(3y+2)-5y&=4\\
6y+4-5y&=4\\
y&=0
\end{split}
\end{equation*}
\begin{SCfigure}
\includegraphics[width=7cm,height=7cm]{exer3b}
\caption{Como se ve en el gr\'afico, correspondinte al punto 3.b, ambas rectas se cortan en el punto $(2,0)$ y por lo tanto, el sistema es compatible determinado.}
\end{SCfigure}
\subsection*{4.a}
\begin{equation*}
\begin{split}
h_{(x)}=f_{(g(x))}&=\sqrt{x^2-2}\\
\\
f_{(x)}&=\sqrt{x}\\
g_{(x)}&=x^2-2
\end{split}
\end{equation*}
\subsection*{4.b}
\begin{equation*}
\begin{split}
h_{(x)}=f_{(g(x))}&=\frac{x+1}{(x+1)^2+2}\\
\\
f_{(x)}&=\frac{x}{x^2+2}\\
g_{(x)}&=x+1
\end{split}
\end{equation*}
\subsection*{5.a}
\begin{equation*}
\begin{aligned}
g_{(x)}=3\cos\left(\frac{x}{2}\right)
\end{aligned}
\end{equation*}
\begin{tabular}{ l | r }
$x$ & $y$\\ \hline
0 & 3\\
1 & 2.63\\
2 & 1.62\\
3 & 0.21\\
$\pi$ & 0\\
4 & -1.25\\
5 & -2.4\\
6 & -2.97\\
$2\pi$ & -3
\end{tabular}
\begin{SCfigure}
\includegraphics[width=7.5cm,height=7.5cm]{exer5a}
\caption{Gr\'afica de la funci\'on $g_{(x)}$, entre $0$ y $2\pi$.}
\end{SCfigure}
\subsection*{5.b}
\begin{equation*}
\begin{aligned}
D=\{0,2\pi\} & \qquad I=\{3,-3\}
\end{aligned}
\end{equation*}
\subsection*{5.c}
\text{La funci\'on no es biyectiva $\mathbb{R}\rightarrow\mathbb{R}$ dado que, por ejemplo, para $x=3\pi\rightarrow y=0$ o para $x=4\pi\rightarrow y=3$, etc.} \\
{Sin embargo, s\'i lo es dentro de $D_f\rightarrow I_f$.}
\subsection*{6.a}
\begin{equation*}
\begin{aligned}
f_x=\left\{
\begin{array}{l l}
-2 & \quad -1\leq x < 0 \\
3-x & \quad 0\leq x \leq \\
-\frac{x^2}{2}+8 & \quad 3< x \leq 5
\end{array}
\right.
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
\text{La siguiente f\'ormula nos indica cu\'al es el eje de simetr\'ia de la funci\'on:}\\
x_v&=\frac{-b}{2a}=\frac{0}{2 \frac{1}{2}}=0\\
\text{Reemplazando $x_v$ en la funci\'on, obtenemos que...}\\
y_v&=-\frac{0^2}{2}+8=8\\
\text{...de lo cual obtenemos el v\'ertice de la par\'abola.}\\
v&=(x_v,y_v)=(0,8)
\end{aligned}
\end{equation*}
\begin{SCfigure}
\includegraphics[width=7.5cm,height=7.5cm]{exer6a}
\caption{Gr\'afica de la funci\'on $f_{(x)}$, del ejercicio 6.}
\label{3.a}
\end{SCfigure}
\end{document}
答案1
您可以使用minipage
s 来解决这个问题,如下所示
\documentclass{article}
\usepackage[demo]{graphicx}
\usepackage{sidecap}
\begin{document}
\begin{SCfigure}
\begin{minipage}{.5\textwidth}
\begin{equation}
f(x)=x^2
\end{equation}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\includegraphics{myimage}
\caption{My figure}
\end{minipage}
\end{SCfigure}
\end{document}
采用minipage
一个强制性参数,,<width>
我已经根据指定了\textwidth
;您可以根据需要进行更改。
答案2
@cmhughes 很完美,但请注意,您可以简单地使用figure
而不是SCfigure
:
\begin{figure}
\begin{minipage}{.5\textwidth}
\begin{equation}
f(x)=x^2
\end{equation}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\includegraphics{myimage}
\caption{My figure}
\end{minipage}
\end{figure}
这避免了额外的依赖。