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\begin{document}
\begin{tabular}{p{3in}|p{1in}}
\ul{APPROXIMATIONS (CONT'D)}
$T'=T\left(1-\frac{v^2}{c^2}\right)^{1/2}\approx T\left(1+\frac 12\frac{v^2}{c^2}\right)$
$\frac{\Delta T}{T}=\frac{1}{2}\frac{v^2}{c^2}\phantom{\approx}\Delta T=T'-T$
&
Error fraction

is proportional

to $\frac{v^2}{c^2}$

with factor $\frac 12$.

2BII1: $\Delta v\approx a\Delta h$
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\ul{QUADRATIC APPROX}

(Use these when linear is not enough)

$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2$

$xneq 0$

&

why $\frac 12 f''(0)$?

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$f(x)=a+bx+cx^2$

$f'(x)=b+2cx$

$f''(x)=2c$


&
recover $a,b,c$

$f(0)=a$

$f'(0)=b$

$f''(0)=2c$




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$\sin x\approx x$

$\cos x\approx 1-\frac 12 x^2$

$e^x\approx 1+x+\frac 12 x^2$

($x$ near 0)

&

$\ln(1+x)\approx x-\frac 12 x^2$

$(1+x)^r\approx 1+rx+\frac{r(r-1)}2x^2$



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\begin{tabular}{p{2.5in}|p{2.5in}}

$a_k\left(1+\frac 1k\right)^k\approx e_{k\rightarrow \infty}$

as $k\rightarrow\infty$:

$\ln a_k=k(\ln(1+\frac{1}{k}))\rightarrow 1$

$\phantom{\ln a_k}\approx k\left(\frac{1}{k}\right)$

$\ln(1+x)\approx x (x=\frac 1k\approx 0)$

&

rate of convergence

$\underbrace{(\ln a_k)}_{text{how big?}}-1\rightarrow 0$

use quadratic approximation.



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\begin{tabular}{l}
Find quadratic approx.\\

$x\approx 0$\\

$e^{-3x}\left(1+x\right)^{-1/2}\approx\left(1+\left(-3x\right)+\frac{\left(-3x\right)^2}2\right)(1-\frac 12 x + \frac 12\underbrace{\left(\frac{-1}{2}\right)}_{r}\underbrace{\left(\frac{-3}{2}\right)}_{r-1}x^2)$\\

$\phantom{e^{-3x}\left(1+x\right)^{-1/2}}\approx 1-\underbrace{3x}_{\frac{1}{100}}-\underbrace{\frac 12 x}_{\frac{1}{100}} +\underbrace{\frac 32 x^2}_{\frac{1}{(100)^2}}+\frac 92 x^2-\frac 38 x^2$\\

$\phantom{e^{-3x}\left(1+x\right)^{-1/2}}=1-\frac 72 x + \frac{51}{8}x^2$\\

(drop $x^3,x^4$, etc. terms of size about $\frac{1}{100^3}$)

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\begin{tabular}{p{1in}p{1in}|p{1in}p{1in}}

$f$&at $x=0$&$f$&at $x=0$\\

$\ln(1+x)$&0&$(1+x)^r$&1\\

$f'=\frac{1}{1+x}$&1&$r(1+x)^{r-1}$&$r$\\

$f''=-\frac{1}{(1+x)}^2$&-1&$r(r-1)(1+x)^{r-2}$&$r(r-1)$



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\ul{CURVE SKETCHING}



GOAL: Draw graph of $f$



using $f'$,$f''$ positive/negative

&

\ul{WARNING:}



Don't abandon your 



precalc skills!



common sense!



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\begin{tabular}{p{2in}|p{2in}}

$f'>0\rightarrow f$ is increasing

\begin{asy}

size(2inch);

import graph;

real f(real x){

return (x*x+2*x+3)/2;

}

draw(graph(f,1,2));

draw((1,f(1.5)-1.25)--(2,f(1.5)+1.25),red);

\end{asy}

&

$f''>0\rightarrow f'$ is increasing

\begin{asy}

size(2inch,2inch);

import graph;

real f(real x){

return (x*x+2*x+3)/2;

}

draw(graph(f,-3,-1));

draw((-3,f(-2.5)+.75)--(-1.75,f(-2.5)-.375*3),red);

draw((-1,f(-1.5)-.25)--(-2.25,f(-1.5)+.375*1),red);

\end{asy}

$f$ concave up

$f''<0\rightarrow f$ concave down



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\ul{Ex 1}. $f(x)=3x-x^3$

$f'(x)=3-3x^2$

$\phantom{f'(x)}=3(1-x)(1+x)$



$f'(x)=0\rightarrow (1-x)(1+x)=0\rightarrow x=\pm 1$

&

$-1<x<1\rightarrow f'(x)>0$

$f$ increasing.

$1<x\rightarrow f'(x)<0$

$f$ decreasing.

$x<-1$ also decreasing.

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schematic:

\begin{asy}

size(2inch,1inch);

draw((-2,0)--(2,0));

draw((-2,1)--(-1,.25)--(1,1)--(2,.25));

dot((-1,.25),2+red);

dot((1,1),2+red);

label("-1",(-1,0),S);

label("1",(1,0),S);

draw((-1,.1)--(-1,-.1));

draw((1,.1)--(1,-.1));

label("turning points",(-2,-1.25),red);

\end{asy}

&

\ul{Defn} If $f'(x_0)=0$ we call

$x_0$ a \textit{critical point}

$y_0=f(x_0)$ is a \textit{critical value}

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\begin{tabular}{p{2in}|p{2in}}
Plot critpts/vals

$f(1)=3\cdot 1-1^3=2$

$f(-1)=3(-1)-(-1)^3=-2$

$f(x)=3x-x^3$ ($f(0)=0$)

&

\begin{asy}
size(1.75inch);
import graph;

dot((-1,-2),L="(-1,-2)",S);

dot((1,2),L="(1,2)",N);

xtick((-1,0));

xtick((1,0));

real f(real x){

return 3*x-x^3;

}

draw(graph(f,-1.1,-.9));

draw(graph(f,1.1,.9));

dot((0,0));

draw(graph(f,-2,-1.1),dotted);

draw(graph(f,-.9,.9),dotted);

draw(graph(f,1.1,2),dotted);

axes();

\end{asy}

$f$ is odd. $f(0)=0$



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\ul{Ends} $x\rightarrow +\infty$

$f(x)=3x-x^3\approx x^3\rightarrow -\infty$

$\phantom{f(x)=3x-x^3\approx} x\rightarrow +\infty$

$f(x)\rightarrow +\infty$ if $x\rightarrow -\infty$

&

$f''(x)=(3-3x^2)'=-6x$

$f''(x)<0$ if $x>0$ (concave down)

$f''(x)>0$ if $x<0$ (concave up)

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\begin{tabular}{p{2in}|p{2in}}
Plot critpts/vals

$f(1)=3\cdot 1-1^3=2$

$f(-1)=3(-1)-(-1)^3=-2$

$f(x)=3x-x^3$ ($f(0)=0$)

&
\begin{asy}
size(1.75inch);
import graph;

dot((-1,-2),L="(-1,-2)",S);

dot((1,2),L="(1,2)",N);

xtick((-1,0));

xtick((1,0));

real f(real x){

return 3*x-x^3;

}

draw(graph(f,-1.1,-.9));

draw(graph(f,1.1,.9));

dot((0,0),red);

label(minipage("inflection point:$f''(0)=0$"),(0,0),SW);

draw(graph(f,-2,-1.1),dotted);

draw(graph(f,-.9,.9),dotted);

draw(graph(f,1.1,2),dotted);

axes();

\end{asy}

$f$ is odd. $f(0)=0$

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