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\title[]{MA140-Engineering Calculus}
\author[ Lecture 14 ]{\large{ Lecture 14} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}
\begin{frame}
\titlepage
\end{frame}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\begin{example}
sketch a graph of the function $f(x)=x^{4}-4x^{3}+10$ 
\end{example}
We use the following steps:\\
\begin{itemize}
\item[(1)] Find the critical (stationary) points
\item[(2)] Find the points of inflection
\item[(3)] Use the second derivative test 
\item[(4)] Find the y-value of these points to pair them
\item[(5)] Draw the table to find the intervals on which $f$ is increasing and the intervals on which $f$ is decreasing
\item[(6)] Add some extra rows to your table to see where the graph of $f$ is concave up and where it is concave down 
\item[(7)] Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. 
\item[(8)] Sketch the general shape of the graph for $f$

\end{itemize}
\end{frame}
\begin{frame}

{\textcolor{red}{Step $(1)$}}:\\
$$f'(x)=4x^{3}-12x^{2}=4x^{2}(x-3)=0$$
So the stationary points are $x=0$ and $x=3$\\
{\textcolor{red}{Step $(2)$}}:\\
$$f''(x)=12x^{2}-24x=12x(x-2)=0$$
So the points of inflection are $x=0$ and $x=2$\\
{\textcolor{red}{Step $(3)$}}:\\
At $x=0$, $f''(0)=0$ so the test fails in this point. But at $x=3$, $f''(3)=36>0$ so based on the test, $x=3$ is a local minimum\\
{\textcolor{red}{Step $4$}}:\\
$f(0)=10$, $f(2)=-6$ and $f(3)=-17$
 
\end{frame}

\begin{frame}

{\textcolor{red}{Step $5$ and $6$}}:
\begin{center}
			\includegraphics[scale=0.3]{Figures/lec111.pdf}
		\end{center}
\end{frame}
\begin{frame}

{\textcolor{red}{Step $7$ and $8$}}:
\begin{center}
			\includegraphics[scale=0.3]{Figures/lec112.pdf}
		\end{center}
\end{frame}
\begin{frame}


\begin{example}
Sketch the graph of $$f(x)=\frac{(x+1)^{2}}{1+x^{2}}$$
\end{example}
{\textcolor{red}{Step $(1)$}}:\\
$\displaystyle{f(x)=\frac{(x+1)^{2}}{1+x^{2}}}$, so $$f'(x)=\frac{(1+x^{2})\cdot 2(x+1)-(x+1)^{2}\cdot 2x}{(1+x^{2})^{2}}=\frac{2(1-x^{2})}{(1+x^{2})^{2}}$$
So the critical points are $x=-1$ and $x=1$. \\
{\textcolor{red}{Step $(2)$}}:\\
$$f''(x)=\frac{(1+x^{2})^{2}\cdot 2(-2x)-2(1-x^{2})[2(1+x^{2})\cdot 2x]}{(1+x^{2})^{4}}=\frac{4x(x^{2}-3)}{(1+x^{2})^{3}}$$
\end{frame}
\begin{frame}
So inflection points are $x=\sqrt{3}$, $x=0$ and $x=-\sqrt{3}$\\
{\textcolor{red}{Step $(3)$}}:\\
since $f''(1)=-1<0$ so $x=1$ is a local max and since $f''(-1)=1>0$ so $x=-1$ is a local min\\
{\textcolor{red}{Step $(4)$}}:\\
$f(1)=2$, $f(-1)=0$, $f(\sqrt{3})=\frac{(\sqrt{3}+1)^{2}}{4}$, $f(-\sqrt{3})=\frac{(-\sqrt{3}+1)^{2}}{4}, f(0)=1$\\
We can also find the asymptotes of the function: $$\lim_{x\to \infty}f(x)=1$$
So $y=1$ is the horizontal asymptote.\\
\end{frame}
\begin{frame}

{\textcolor{red}{Step $5$ and $6$}}:
\begin{center}
			\includegraphics[scale=0.3]{Figures/lec130.pdf}
		\end{center}
\end{frame}
\begin{frame}

{\textcolor{red}{Step $7$ and $8$}}:
\begin{center}
			\includegraphics[scale=0.3]{Figures/lex131.pdf}
		\end{center}
\end{frame}
\begin{frame}{Exercise }
Use the steps of the graphing procedure to graph the following equations:
\begin{itemize}
\item $$y=x^{2}-4x+3$$
\item $$y=x^{3}-3x+3$$
\item $$y=\frac{x^{2}-3}{x-2}$$
\end{itemize}

\end{frame}







\end{document}