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\documentclass[12pt, letterpaper]{report}
\usepackage[utf8]{inputenc}
\usepackage[legalpaper, portrait, top=0.0in, bottom=-0.8in, left=0.9in, right=0.9in]{geometry}
\usepackage{titlesec}
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\usepackage{graphicx}
\graphicspath{ {images/} }
\usepackage{setspace}
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\usepackage{tikz}
\usetikzlibrary{datavisualization.formats.functions}
\usepackage{pgfplots}
\pgfplotsset{my style/.append style={axis x line=middle, axis y line=
middle, xlabel={$x$}, ylabel={$y$}, axis equal }}
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\setlength{\parskip}{1em}
\begin{document}
\chapter{}
\section{Secants and Tangents}
\begin{flushleft}
If we consider function $f(x)$ and graph it on a Cartesian plane, different points on the curve are going to be more 'steep' than others, as evident through the graph below. By looking closely at the two points, it can be clearly shown that despite both points existing on the same curve, $f(x)$ at Point A is much more steep than Point B.
\\
\begin{center}
\begin{tikzpicture}
\begin{axis}[
x=1.0cm,y=1.0cm,
axis lines=middle,
xmin=-1.5,
xmax=6.5,
ymin=-1.0,
ymax=4.0298,
xtick={-1.0,0.0,...,6.0},
ytick={-1.0,0.0,...,4.0},]
\clip(-1.5,-1.) rectangle (6.5,4.0298);
\draw[line width=1.pt,smooth,samples=100,domain=-0.9999920000000018:6.5] plot(\x,{2*ln((\x)+1)});
\begin{scriptsize}
\draw[color=black] (-0.2,-0.93) node {$f$};
\draw [fill=black] (0.94,1.3253759461504735) circle (2.5pt);
\draw[color=black] (1.08,1.69) node {$A$};
\draw [fill=black] (4.76,3.5018749494156) circle (2.5pt);
\draw[color=black] (4.9,3.87) node {$B$};
\end{scriptsize}
\end{axis}
\end{tikzpicture}
\\
\fontsize{8pt}{8pt}Figure 1
\\
\end{center}
\par
Mathematicians are able to determine how steep a curve is at a point by finding the \emph{gradient}. However, we haven't dealt with changing gradients before, as linear equations in the form $y=mx+b$ had a constant gradient of $m$, which is easily found.
\par
As such, we can find the gradient at various points by constructing a line which only touches the curve at each respective point.
\\
\begin{center}
\begin{tikzpicture}
\begin{axis}[
x=1.0cm,y=1.0cm,
axis lines=middle,
xmin=-1.5,
xmax=7.0,
ymin=-1.0,
ymax=5.0,
xtick={-0.0,1.0,2.0,...,6.0},
ytick={-0.0,1.0,2.0,...,4.0},]
\draw[line width=1.pt,smooth,samples=100,domain=-0.8:7.0] plot(\x,{2*ln((\x)+1)});
\draw [line width=1.pt,domain=-1.5:7.] plot(\x,{(--0.39--1.*\x)/1.});
\draw [line width=1.pt,domain=-1.5:7.] plot(\x,{(--1.8878--0.34*\x)/1.});
\begin{scriptsize}
\draw[color=black] (-0.6569985630725312,-5.94725801699136) node {$f$};
\draw [fill=black] (1.,1.3862943611198906) circle (2.5pt);
\draw[color=black] (1.1645293126567067,1.8292648370833955) node {$A$};
\draw [fill=black] (4.831478851964947,3.526541261250383) circle (2.5pt);
\draw[color=black] (4.994408435984848,3.9543806921008215) node {$B$};
\draw[color=black] (8.917699245247823,9.699199377092983) node {$g$};
\draw[color=black] (-7.4994045577990285,-0.24914517496661417) node {$h$};
\end{scriptsize}
\end{axis}
\end{tikzpicture}
\\
\fontsize{8pt}{8pt}Figure 2
\end{center}
So how can we find the gradient of these tangent? For more curves, its inaccurate to just approximate the curve however we can actually use \emph{limits} to find our tangent.
\newpage
\fbox{\begin{minipage}{40em}
\begin{center}
\textbf{Revision:Limits}
\end{center}
\end{minipage}}
\end{flushleft}
\end{document}