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\title[]{MA140-Engineering Calculus}
\author[ Lecture 13 ]{\large{ Lecture 13} 
\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{definition}
The graph of a differentiable function $y=f(x)$ is:
\begin{itemize}
\item \textbf{concave up} on an open interval $I$ if $f'$ is increasing on $I$
\item \textbf{concave down} on an open interval $I$ if $f'$ is decreasing on $I$
\end{itemize}
\end{definition}
\begin{theorem}
Let $y=f(x)$ be twice-differentiable on an open interval $I$.
\begin{itemize}
\item If $f''>0$ on $I$, the graph of $f$ over $I$ is concave up
\item If $f''<0$ on $I$, the graph of $f$ over $I$ is concave down. 
\end{itemize}
\end{theorem}
\end{frame}
\begin{frame}
\begin{center}
			\includegraphics[scale=0.2]{Figures/lec99.pdf}
		\end{center}
		\begin{center}
			\includegraphics[scale=0.18]{Figures/lec100.pdf}
		\end{center}
\end{frame}
\begin{frame}
\begin{definition}
A point where the graph of a function has a tangent line and where the concavity changes is a \textbf{point of inflection}
\end{definition}
{\textcolor {red} {\textit{Note:}}} To find the points of inflection, we need to find the zeros  of $f''$ and points where $f''$ is not defined. Then we need to check the concavity.
\begin{example}
The curve $y=x^{4}$ has no inflection point at $x=0$. Even though $y''=12x^{2}$ is zero there, it does not change sign
\end{example}
{\textcolor {red} {\textit{Note:}}} An inflection point may not exist where $y''=0$
\end{frame}
\begin{frame}
\begin{center}
			\includegraphics[scale=0.2]{Figures/lec113.pdf}
		\end{center}
		\begin{theorem}
		{\textcolor{red}{Second derivative test for local maximums and minimums:}} suppose that $f''$ is continuous on an open interval that contains $x=c$.
		\begin{itemize}
		\item If $f'(c)=0$ and $f''(c)<0$, then $f$ has a local maximum at $x=c$
		\item If $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $x=c$
		\item If $f'(c)=0$ and $f''(c)=0$, then the test fails. The function $f$ may have a local maximum, a local minimum, or neither.
\end{itemize}		 
		\end{theorem}
\end{frame}
\begin{frame}
\begin{example}
Find and classify the stationary points and the points of inflection of $$f(x)=4x^{3}-21x^{2}+18x+6$$
\end{example}
$f'(x)=12x^{2}-42x+18$\\
When $f'(x)=0$, we have: $$12x^{2}-42x+18=0 \Rightarrow 2x^{2}-7x+3=0 \Rightarrow (2x-1)(x-3)=0$$
So The stationary points are: $x=1/2$ and $x=3$\\
$f''(x)=24x-42$ so $$f''(1/2)=24(1/2)-42=12-42<0$$
which means $x=1/2$ is a local max. Also as $$f''(3)=24(3)-42=72-42>0$$
$x=3$ is a local min. 
\end{frame}












\end{document}