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practicefinal2023.js
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practicefinal2023.js
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import {
rp,
rpp,
addpause,
sub,
picsize,
cleanpoly,
zerosquadfactor,
convertWork,
deriv,convterm, pow
} from "./Calc.js?v=160";
function fpaex(fs,a){
let str=convterm(fs)
let fps=addpause(deriv(str)[1]);
let str1='f\'(x)&\\p{=}'+fps;
let str2='\\p{f\'(\\r{'+a+'})}&\\p{=}'+sub(deriv(str)[1],'x','\\r{'+a+'}')
return {
question:'<p>Let $f(x)='+rp(fs)+`.$</p><p>Find $f'(\\r{`+a+`})$</p>`,
steps:[
'$$\\begin{aligned}'+str1+'\\end{aligned}$$',
'$$\\begin{aligned}'+str2+'\\end{aligned}$$'
]
}
}
function eval1(fs,a){
return fs.replaceAll('x','{('+a+')}')
}
function pwfn(name,f,g,a,sign1,sign2){
a='\\go{'+a+'}'
if(sign1=='<'||sign1=='\\le'){
f='\\r{'+f+'}';
g='\\g{'+g+'}';
sign1='\\r{'+sign1+'}'
sign2='\\g{'+sign2+'}'
}
else{
f='\\g{'+f+'}';
g='\\r{'+g+'}';
sign1='\\g{'+sign1+'}'
sign2='\\r{'+sign2+'}'
}
let i=8;
if(f.includes('\\frac')){
i=i+3;
}
if(g.includes('\\frac')){
i=i+3;
}
let str=name==''?'':'\\p{'+name+'(x)}\\p{=}'
str=str+'\\p{\\begin{cases}'+rpp(f)+'&\\p{x}\\p{'+sign1+'}\\p{'+a+'}\\\\['+i+'pt]'+rpp(g)+'&\\p{x}\\p{'+sign2+'}\\p{'+a+'}\\end{cases}}'
return str.replaceAll('x','\\b{x}')
}
function sublim(f,a){
return sub(f,'\\b{x}','\\go{'+a+'}')
}
function lim(f,a){
return limx('x',f,a)
}
function liml(f,a){
return limxs('x',f,a,'\\r{-}')
}
function limr(f,a){
return limxs('x',f,a,'\\g{+}')
}
function limxs(x,f,a,s){
let str=s==''?'':'^\\p{'+s+'}';
return '\\p{\\lim\\limits_{\\p{\\b{'+x+'}}\\p{\\to}\\p{\\go{'+a+'}}'+str+'}}'+f
}
function bx(f,x){
return f.replaceAll(x,'\\b{'+x+'}')
}
function pwfnex(f,g,a,sign1,sign2,k1,k2){
let fstr='\\p{f(\\b{x})}'
let qstr='$$'+rpp(pwfn('f',f,g,a,sign1,sign2))+String.raw`$$<span class="invisible">Find each of the following:</span><ol><pli>$`+rp(liml(fstr,a))+`$</pli><pli>$`+rp(limr(fstr,a))+`$</pli><pli>$`+rp(lim(fstr,a))+`$</pli><ol>`;
f=addpause(f);
g=addpause(g);
let f1=f
let g1=g
let n1=k1;
let n2=k2;
if(sign1=='>'||sign1=='\\ge'){
f1=g
g1=f
n1=k2;
n2=k1;
}
f1=bx('\\r{'+f1+'}','x')
g1=bx('\\g{'+g1+'}','x')
let i1=4;
let i2=4;
if(f1.includes('\\frac')){
i1=10;
i2=i2+3;
}
let j1=4;
let j2=4;
if(g1.includes('\\frac')){
j1=10;
j2=j2+3;
}
if(n1 instanceof String &&n1.includes('\\frac')){
i2=i2+3;
}
if(n2 instanceof String &&n2.includes('\\frac')){
j2=j2+3;
}
let str='Since $'+liml(fstr,a)+'\\p{=}'+limr(fstr,a)+'\\p{=}\\p{\\pu{'+n1+'}}$,</p><p>$'+lim(fstr,a)+'\\p{=}\\p{\\pu{'+n1+'}}$';
if(n1!=n2){
str='Since $'+liml(fstr,a)+'\\p{\\ne}'+limr(fstr,a)+'$</p><p>$'+lim(fstr,a)+'\\p{=}\\p{\\t{DNE}}$';
}
str='What were the $\\rt{left}$ and $\\gt{right}$ limits?\\p$$\\begin{aligned}'+liml(fstr,a)+'&\\p{=}\\p{'+n1+'}\\\\[10pt]'+limr(fstr,a)+'&\\p{=}\\p{'+n2+'}\\end{aligned}$$</p><p>'+str
a='\\go{'+a+'}'
return{
question:qstr,
steps:[
`Since $\\b{x}\\to`+a+`^{\\r{-}}$\\p means that\\p $\\b{x}$ is approaching $`+a+`$\\p and\\p ` +
`<span class='red'>less than</span>\\p $`+a+`$, $`+liml('\\p{f(\\b{x})}',a)+`\\p{=}`+liml(f1,a)+'$$$\\begin{aligned}'+liml(fstr,a)+'&\\p{=}'+liml(f1,a)+'\\\\['+i1+'pt]&\\p{=}'+sublim(f1,a)+'\\\\['+i2+'pt]&\\p{=}\\p{'+n1+'}\\end{aligned}$$',
'Since $\\b{x}\\to'+a+'^{\\g{+}}$\\p means that\\p $\\b{x}$ is approaching $'+a+'$\\p and\\p<span' +
' class="green"> greater than</span>\\p $'+a+'$, $'+limr('\\p{f(\\b{x})}',a)+'\\p{=}'+limr(g1,a)+'$$$\\begin{aligned}'+limr(fstr,a)+'&\\p{=}'+limr(g1,a)+'\\\\['+j1+'pt]&\\p{=}'+sublim(g1,a)+'\\\\['+j2+'pt]&\\p{=}\\p{'+n2+'}\\end{aligned}$$',
str
]
}
}
function incdecsetup(a1,x1,x2,d){
let temp1=x1;
x1=Math.min(x1,x2);
x2=Math.max(temp1,x2)
let list=cubicsetup(a1,x1,x2,d);
return incdec(list[0],list[1],list[2],list[3],list[4])
}
function incdec(a,b,c,d,zs){
return signchartprob(a,b,c,d,zs,'','increasing','decreasing');
}
function cubicsetup(a1,x1,x2,d){//,bincdec
let a=a1;
let b=-3*a*(x1+x2);
let c=3*a*x1*x2;
if(b%2==0){
b=b/2;
}
else if(a%2!=0){
a=a*2;
c=c*2;
}
return [a,b,c,d,makezs(x1==x2?[x1]:[x1,x2])];//Doesn't work for cubic but cubic generates its own zs
}
function as(list){
return list.length>1?'s':''
}
function xss(xs){
let xssol='x=';
for(let i=0;i<xs.length;i++){
let x=xs[i];
xssol=xssol+String.raw`\o{`+x+String.raw`}`;
if(i<xs.length-1){
xssol=xssol+",\\ ";
}
}
return xssol;
}
function posnegzs(f,xs,zs){
let n=zs.length-1;
let pzsi=[];
let nzsi=[];
let str='\\p{(}\\p{-}\\p{\\infty}\\p{,}\\p{\\o{'+xs[0]+'}}\\p{)}';
if(f(zs[0])>=0){
pzsi=pzsi.concat(str);
}
else{
nzsi=nzsi.concat(str);
}
for(let i=1;i<zs.length-1;i++){
str='\\p{(}\\p{\\o{'+xs[i-1]+'}}\\p{,}\\p{\\o{'+xs[i]+'}}\\p{)}';
if(f(zs[i])>=0){
pzsi=pzsi.concat(str);
}
else{
nzsi=nzsi.concat(str);
}
}
str='\\p{(}\\p{\\o{'+xs[n-1]+'}}\\p{,}\\p{\\infty}\\p{)}';
if(f(zs[n])>=0){
pzsi=pzsi.concat(str);
}
else{
nzsi=nzsi.concat(str);
}
let v=[];
v.push(pzsi,nzsi);
return v;
}
function fsignex1(fn,szsi,ineq,fps1,ineq11){
let fpexp1=`<pli>$`+fn+`(x)`;
if(szsi.length==0){
if(ineq.length>8){
ineq=ineq.substring(8);
}
return fpexp1+`\\p{\\t{ is}}\\p{\\t{ never }`+ineq+`}$</pli>`;
}
fpexp1=fpexp1+ineq+'\\t{ on }';
for(let i=0;i<szsi.length;i++){
fpexp1=fpexp1+szsi[i];
if(i<szsi.length-1){
fpexp1=fpexp1+'\\p{,}\\ '
}
}
if(fps1.length>0){
fpexp1=fpexp1+'\\p{\\t{ b/c }}\\p{'+fps1+'(x)}\\p{'+ineq11+'}';
}
return fpexp1+'$</pli>';
}
function list2str(list){
let str='';
for(let i=0;i<list.length;i++){
str=str+(i>0?'\\p{,} ':'')+'\\p{\\o{'+list[i]+'}}'
}
return str
}
function fsignex(fn,pzsi,nzsi,ineq1,ineq2,fps1){
return '<p><ol>'+fsignex1(fn,pzsi,ineq1,fps1,'\\g{>0}')+fsignex1(fn,nzsi,ineq2,fps1,'\\r{<0}')+'</ol></p>';
}
function signcharts(f,xs,zs){
let signs=[]
let strs=[];
for(let i=0;i<zs.length;i++){
signs[i]=Math.sign(f(zs[i]))
}
for(let i=0;i<3;i++){
strs[i]=createSignChartCanvas(xs,zs,signs,true,i>0,i>1,i)//signchartfig(f,xs,zs,strs1[i]);
}
return strs;
}
function drawRec(content,x,y,w,h,c){
content.fillStyle = "rgb("+c[0]+","+c[1]+","+c[2]+")"
content.fillRect(x-w/2,y-h/2,w,h);
}
function drawText(content,text,x,y,c,fontsize){
if(fontsize==null){
fontsize=30;
}
content.font = fontsize+"px Arial";
content.textAlign = "center";
content.fillStyle = "rgb("+c[0]+","+c[1]+","+c[2]+")"
content.fillText(text,x,y);
}
function signchartprob(a,b,c,d,zs,fps1,qstr1,qstr2){
let f=cleanpoly(a,b,c,d)
let fp=cleanpoly(0,3*a,2*b,c);
let fpfunc=x=>3*a*Math.pow(x,2)+2*b*x+c;
let sol=fps1==''?zerosquadfactor(3*a,2*b,c):zerosquadfactor(0,6*a,2*b)
let zw=convertWork(sol[0])
let xs=sol[1]
xs.sort();
let xssol=xss(xs);
let str0='';
let fn=`f'`;
let strs1=['\\t{ is }\\gt{increasing}','\\t{ is }\\rt{decreasing}',`f'`]
if(fps1.length>0){
str0=`f'(x)&\\p{=}`+addpause(fp)+'\\\\[4pt]\\p'
fpfunc=x=>6*a*x+2*b;
fp=cleanpoly(0,0,6*a,2*b)
fn=`f''`;
strs1=['\\t{ is concave }\\gt{up}','\\t{ is concave }\\rt{down}',`f''`];
}
let strs=signcharts(fpfunc,xs,zs);
//let critstr=String.raw`<p>\p$$\gb{`+fp+String.raw`=0}$$`;
// if(zw.length>0){
// critstr=critstr+String.raw`\p$$\bb{\a{`+zw+`}}$$`
// }
// critstr=critstr+`\\p$$\\gb{`+xssol+`}$$</p>`;
let posneg=posnegzs(fpfunc,xs,zs);
let pzsi=posneg[0];
let nzsi=posneg[1];
return {question: String.raw`<p>$$f(x)=`+f+`$$Find the intervals where $f(x)$ is `+qstr1+` and where $f(x)$ is `+qstr2+`</p>`,
steps: [
`<p>$$\\begin{aligned}`+str0+`{`+fn+`(x)}&\\p{=}`+addpause(fp)+`\\end{aligned}$$</p>`,
zw,
strs[0],//pic(strs)
strs[1],
'$$\\begin{aligned}'+fn+'(x)&\\p{=}\\p{'+fp+'}\\\\[8pt]'+fevalsigns(fpfunc,fp,fn,zs)+'\\end{aligned}$$',
strs[2],
fsignex(fn,pzsi,nzsi,'\\g{>0}','\\r{<0}',''),
fsignex('f',pzsi,nzsi,strs1[0],strs1[1],strs1[2]),
],
}
}
function fevalsigns(f,fs,fn,zs){
let str='';//'$$\\begin{aligned}';
for(let i=0;i<zs.length;i++){
let r='\\p{(\\pu{'+zs[i]+'})}';
let y=f(zs[i]);
let ineq='\\g{>0}';
if(y<0){
ineq='\\r{<0}';
}
str=str+(i>0?'\\\\[16pt]':'')+'\\p{'+fn+r+'}&\\p{=}'+sub(addpause(fs),'x','\\pu{'+zs[i]+'}')+'\\\\[4pt]&\\p{=}\\p{'+y+'}\\p{'+ineq+'}'
// if(i<zs.length-1){
// str=str+'\\\\[4pt]';
// }
}
return str//+'\\end{aligned}$$';//+'}$$</p>'
}
function makezs(xs) {
let n=xs.length;
if (xs.length == 0) {
return [0];
}
let zs = [];
zs[0]=xs[0]>0?0:xs[0]-1;
for(let i=1;i<n;i++){
let x1=xs[i-1]+1;
let x2=xs[i]-1;
if(xs[i-1]<0&&xs[i]>0){
zs[i]=0;
}
else {
zs[i]=Math.abs(x1)>Math.abs(x2)?x2:x1;
}
}
zs[n]=xs[n-1]<0?0:xs[n-1]+1;
return zs;
}
let names=[];
let xssc=[];
let zssc=[];
let signssc=[];
let bxssc=[];
let bzssc=[];
let bsignssc=[];
function drawSignChart(name,xs,zs,signs,bxs,bzs,bsigns){
let canvas = document.getElementById(name)
let content=canvas.getContext("2d");
let y0=200;
let M=xs.length;
let a=canvas.width;
let dx=8;
let dycrit=54;
let dytest=27;
let xaxish=4;
let critColor=[253, 119, 0];
let testColor=[153, 102, 255];
let deltay=35;
let deltay2=50;
let dyplus=40;
let plusColor=[114, 221, 0];
let minusColor=[232, 9, 0];
drawRec(content,a/2,y0,a,xaxish,[0,0,0])
for(let i=0;i<M+1;i++){
if(i<M&&bxs){
let x=a/(M+1)*(i+1)
drawRec(content,x,y0,dx,2*dycrit,critColor)
drawText(content,xs[i],x-(xs[i]<0?4:0),y0+dycrit+deltay,critColor)
}
let x=a/(M+1)*(i+0.5);
if(bsigns&&i<signs.length){
if(signs[i]>0){
drawRec(content,x,y0-dytest-deltay2-dyplus,2*dyplus,dx,plusColor)
drawRec(content,x,y0-dytest-deltay2-dyplus,dx,2*dyplus,plusColor)
}
if(signs[i]<0){
drawRec(content,x,y0-dytest-deltay2-dyplus,2*dyplus,dx,minusColor)
}
}
if(bzs&&i<zs.length){
drawRec(content,x,y0,dx,2*dytest,testColor)
drawText(content,zs[i],x-(zs[i]<0?4:0),y0+dytest+deltay,testColor)
}
}
}
document.addEventListener("DOMContentLoaded", function(event) {
for(let i=0;i<names.length;i++){
drawSignChart(names[i],xssc[i],zssc[i],signssc[i],bxssc[i],bzssc[i],bsignssc[i])
}
});
function btotext(b){
return b?'T':'F';
}
function signchartName(xs,zs,signs,bxs,bzs,bsigns){
let str='SignChart_xs_';
for(let i=0;i<xs.length;i++){
str=str+'_'+xs[i];
}
str=str+'_zs_'
for(let i=0;i<zs.length;i++){
str=str+'_'+zs[i]
}
str=str+'_signs_'
for(let i=0;i<signs.length;i++){
str=str+'_'+(signs[i]>0?'p':'m');
}
let name=str+'_'+btotext(bxs)+'_'+btotext(bzs)+'_'+btotext(bsigns)
let count=1;
while(names.includes(name+'_'+count)){
count=count+1;
}
names.push(name+'_'+count);
return name+'_'+count;
}
function createSignChartCanvas(xs,zs,signs,bxs,bzs,bsigns,j){
let name=signchartName(xs,zs,signs,bxs,bzs,bsigns);
xssc.push(xs);
zssc.push(zs);
signssc.push(signs);
bxssc.push(bxs);
bzssc.push(bzs);
bsignssc.push(bsigns);
return '<canvas id=' + name + ' width="660" height="300" class="invisible">Your browser does not support the canvas element.</canvas>'//style="position:absolute;top:0;left:0;z-index:'+10*j+'"
}
function fenceQ(d){
return `A farmer wants to build a three-sided rectangular fence near a river, using $\\yk{`+d+String.raw`}$ yards of fencing. Assume that the river runs straight and that John need not fence in the side next to the river. What are the $\ot{width}$ and the $\yut{length}$ of the fence which maximizes the enclosed area?`
}
function fence(d){
return {
question:fenceQ(d),
steps:[
'$A\\p{=}\\p{\\pu{x}}\\p{\\cdot}\\p{\\o{y}}$',
String.raw`<p>Based on the below diagram,\p the amount of fence used\p is $$\p{2}\p{\o{x}}\p{+}\p{\yu{y}}$$</p><p><iframe src="https://www.desmos.com/calculator/tkj3ktocc4?embed" width="500" height="500" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>Since the farmer uses\p $\yk{`+d+String.raw`}$\p yards of the fence,\p$$\p{2}\p{\o{x}}\p{+}\p{\yu{y}}\p{=}\p{\yk{`+d+`}}$$</p>`,
String.raw`$$\yu{y}\p{=}\p{\yk{`+d+`}}\\p{-}\\p{2}\\p{\\o{x}}$$`,
String.raw`$$\begin{aligned}A&\p{=}\p{\o{x}}\p{\cdot}\p{\yu{y}}\\[4pt]&\p{=}\p{\o{x}}\p{\cdot}\p{(\yk{`+d+String.raw`}\p{-}\p{2}\p{\o{x}})}\\[4pt]\p{f(\o{x})}&\p{=}\p{\o{x}}\p{\cdot}\p{(\yk{`+d+String.raw`}\p{-}\p{2}\p{\o{x}})}\end{aligned}$$`,
String.raw`$$\begin{aligned}f(\o{x})&\p{=}\p{\o{x}}\p{\cdot}\p{(\yk{`+d+String.raw`}\p{-}\p{2}\p{\o{x}})}\\[4pt]&\p{=}\p{\yk{`+d+String.raw`}}\p{\o{x}}\p{-}\p{2}\p{\o{x}}^\p{2}\\[4pt]\p{f'(\o{x})}&\p{=}\p{\yk{`+d+`}}\\p{-}\\p{4}\\p{\\o{x}}\\end{aligned}$$`,
String.raw`$$\begin{aligned}\p{\yk{`+d+String.raw`}}\p{-}\p{4}\p{\o{x}}&\p{=}\p{0}\\[4pt]\p{\yk{`+d+String.raw`}}&\p{=}\p{4}\p{\o{x}}\\[7pt]\p{\o{x}}&\p{=}\p{\frac{\p{\yk{`+d+String.raw`}}}{\p{4}}}\\[7pt]\p{\o{x}}&\p{=}\p{\ot{`+d/4+`}}\\end{aligned}$$`,
String.raw`$\begin{aligned}f'(x)&\p{=}\p{\yk{`+d+String.raw`}}\p{-}\p{4\o{x}}\\[4pt]\p{f''(x)}&\p{=}\p{-4}\end{aligned}$`,
String.raw`Because\p $f''(x)$ is\p always \p$\rt{negative}$,\p a $\ot{width}$\p of\p $\ot{`+d/4+String.raw`}$ gives\p the $\rt{maximum}$ area`,
String.raw`$$\begin{aligned}\yu{y}&\p{=}\p{\yk{`+d+`}}\\p{-}\\p{2}\\p{\\o{x}}\\\\[4pt]&\\p{=}\\p{\\yk{`+d+`}}\\p{-}\\p{2}\\p{(\\o{`+d/4+`})}\\\\[4pt]&\\p{=}\\p{\\yu{`+d/2+`}}\\end{aligned}$$`,
'$$\\begin{aligned}A&\\p{=}\\p{\\pu{x}}\\p{\\cdot}\\p{\\o{y}}\\\\[4pt]&\\p{=}\\p{(\\o{'+d/4+'})}\\p{\\cdot}\\p{(\\pu{'+d/2+'})}\\\\[4pt]&\\p{=}\\p{'+pow(d,2)/8+'}\\end{aligned}$$',
]
}
}
function ca0(qs,ini,k,t,tu,N,gr){
return{
question: '<p>A '+qs+String.raw` is $\lb{`+ini+`}$. The `+grn(gr,N)+` of the `+qs+` is $\\o{`+k+`\\%}$. Find the $\\yut{`+qs+`}$ after $\\g{`+t+`}$ `+tu+`</p>`,
steps:[
cp(k,N),
`$$\\begin{aligned}\\p{\\lbt{starting amount}}&\\p{=}\\p{\\lb{`+ini+`}}\\\\[6pt]\\p{`+grt(N,'')+`}&\\p{=}\\p{\\o{`+k/100+`}}\\\\[6pt]\\p{\\gt{time passed}}&\\p{=}\\p{\\g{`+t+`}}\\end{aligned}$$$$\\p{`+ca(ini,k/100,t,0,N)+`}$$`,
]
}
}
function ca0g(qs,ini,k,t,tu,gr){
return ca0(qs,ini,k,t,tu,false,gr);
}
function ca0d(qs,ini,k,t,tu,gr){
return ca0(qs,ini,k,t,tu,true,gr);
}
function ca1(qs,cam,k,t,tu,N,gr){
return{
question: '<p>After '+`$\\g{`+t+`}$ `+tu+`, the `+qs+String.raw` is $\yu{`+cam+`}$. The `+grn(gr,N)+` rate of the `+qs+` is $\\o{`+k+`\\%}$. Find the $\\lbt{starting `+qs+`}$</p>`,
steps:[
cp(k,N),
`$$\\begin{aligned}\\p{\\put{current amount}}&\\p{=}\\p{\\pu{`+cam+`}}\\\\[6pt]\\p{`+grt(N,'')+`}&\\p{=}\\p{\\o{`+k/100+`}}\\\\[6pt]\\p{\\gt{time passed}}&\\p{=}\\p{\\g{`+t+`}}\\end{aligned}$$$$\\p{`+ca(cam,k/100,t,1,N)+`}$$`,
String.raw`$$\p{\lbt{starting amount}=}\p{\frac{\p{\yu{`+cam+`}}}{\\p{e}^{`+msp(N)+'\\p{(\\o{'+k/100+`})}\\p{(\\g{`+t+`})}}}}$$`
]
}
}
function grt(N,gr){
return `\\ot{`+grn(gr,N)+` rate}`;
}
function msp(N){
return N?'\\p{\\r{-}}':'';
}
function cp(k,N){
return '$$\\pn{'+grt(N,'')+'=\\p{\\frac{\\p{\\o{'+k+'}}}{\\p{100}}}\\p{=}\\p{\\o{'+k/100+'}}}$$';
}
function grn(gr,N){
if(gr.length==0){
gr=N?'decay':'growth';
}
return gr;
}
function ca(a,b,c,n,N){
let cam=['\\yut{current amount}','\\lbt{starting amount}',grt(N,''),'\\gt{time passed}']
let strs=n==0?['\\pn{','}\\pn{','}']:['','',''];
let is=[0,1,2,3];
is.splice(n,1);
let cs=['pu','lb','o','g'];
let ds=[a,b,c];
for(let i=0;i<is.length;i++){
cam[is[i]]='\\'+cs[is[i]]+'{'+ds[i]+'}';
}
return strs[0]+cam[0]+'='+strs[1]+'\\p{'+cam[1]+'}\\p{\\cdot} \\p{e}^{'+msp(N)+'\\p{('+cam[2]+')}\\p{('+cam[3]+')}}'+strs[2];
}
function ederiv(c){
return '\\ds\\frac{d}{dx}{\\b{'+c+'}e^x}=\\p{\\b{'+c+'}}\\p{e^\\p{x}}'
}
function eaderiv(c,a){
return '\\frac{d}{dx}{\\b{'+c+'}e^{\\r{'+a+'}x}}=\\p{\\r{'+a+'}}\\p{\\cdot}\\p{\\b{'+c+'}}\\p{e^{\\p{\\r{'+a+'}}\\p{x}}}'
}
function clnderiv(c){
return '\\ds\\frac{d}{dx}\\b{'+c+'}\\ln(x)=\\p{\\frac{\\p{\\b{'+c+'}}}{\\p{x}}}'
}
function derivintroex(fs) {
return {
question: '<p>Let $f(x)=' + fs + '$</p><p>Find $f\'(x)$</p>',
steps: ['$$f\'(x)=' + addpause(deriv(fs)[1]) + '$$']
}
}
function consurp(sx,sc,dx,dc){//ss,ds,xe,s
let s=x=>sx*x+sc;
let xe=(dc-sc)/(sx-dx);
let qe='\\go{'+s(xe)+'}'
xe='\\b{'+xe+'}'
let ss=cleanpoly(sx,sc);
let ds=cleanpoly(dx,dc);
ds='\\r{'+ds+'}';
ss='\\g{'+ss+'}';
let xq='\\p{-}\\p{('+xe+')}\\p{('+qe+')}'
return {
question:'Find the consumer surplus for the supply $\\g{S(x)}='+ss+'$, $\\r{D(x)}=\\r{'+ds+'}$, $\\bt{equilibrium quantity}='+xe+'$ and $\\got{equilibrium price}='+qe+'$',
steps:[
'$$\\a{\\p{\\t{consumer surplus}}&\\p{=}\\iep{0}{\\bt{eq. quantity}}{\\r{D(x)}}\\p{-}\\p{(\\bt{eq. quantity})}\\p{(\\got{eq. price})}\\\\[10pt]&\\p{=\\iep{0}{'+xe+'}{'+ds+'}'+xq+'}\\\\[10pt]&\\p{=\\p{\\left('+defintbar(0,xe,ds)+'\\right)}'+xq+'}\\\\[10pt]&\\p{=\\left('+defintplugin(0,xe,ds)+'\\right)'+xq+'}}$$'
]
}
}
function prodsurp(sx,sc,dx,dc){//ss,ds,xe,s
let s=x=>sx*x+sc;
let xe=(dc-sc)/(sx-dx);
let qe='\\go{'+s(xe)+'}'
xe='\\b{'+xe+'}'
let ss=cleanpoly(sx,sc);
let ds=cleanpoly(dx,dc);
xe='\\b{'+xe+'}'
ss='\\g{'+ss+'}'
ds='\\r{'+ds+'}';
let xq='\\p{('+xe+')}\\p{('+qe+')}\\p{-}'
return {
question:'Find the producer surplus for the supply $\\g{S(x)}=\\g{'+ss+'}$, $\\r{D(x)}='+ds+'$, $\\bt{equilibrium quantity}='+xe+'$ and $\\got{equilibrium price}='+qe+'$',
steps:[
'$$\\a{\\p{\\t{producer surplus}}&\\p{=}\\p{(\\bt{eq. quantity})}\\p{(\\got{eq. price})}\\p{-}\\iep{0}{\\bt{eq. quantity}}{\\g{S(x)}}\\\\[10pt]&\\p{='+xq+'\\iep{0}{'+xe+'}{'+ss+'}}\\\\[10pt]&\\p{='+xq+'\\p{\\left('+defintbar(0,xe,ss)+'\\right)}}\\\\[10pt]&\\p{='+xq+'\\p{\\left('+defintplugin(0,xe,ss)+'\\right)}}}$$'
]
}
}
function piw(str){
return '\\p{\\pi\\p{\\left(\\p{'+str+'}\\right)}}'
}
function vol(a,b,fs){//gs
fs='\\g{'+fs+'}'
a='\\r{'+a+'}'
b='\\b{'+b+'}'
return{
question:`Find the volume of the solid generated by revolving $y=`+fs+`$ from $x=`+a+`$ to $x=`+b+`$ around the $x$-axis`,
steps:[
'$$\\a{\\p{\\t{volume}}&\\p{{}=\\p{\\pi}\\iep{\\r{a}}{\\b{b}}{\\p{\\left(\\p{\\g{f(x)}}\\right)}\\p{^2}}}\\\\[10pt]&\\p{=}\\p{\\pi}\\iep{'+a+'}{'+b+'}{\\p{\\left(\\p{'+fs+'}\\right)}\\p{^2}}}$$',
// '$$\\a{\\t{volume}&='+piw(defintstr(a,b,gs))+'\\\\[10pt]&\\p{=}'+piw(defintbar(a,b,gs))+'\\\\[10pt]&\\p{=}'+piw(defintplugin(a,b,gs))+'}$$'
]
}
}
function iterms(fs,is,rs){
let str='$$\\a{\\i{'+fs+'}&='
if(rs.length>0){
str=str+rs+'\\\\[10pt]&\\p{=}'
}
return str+is+'\\C}$$'
}
function itermsprep(fps,fss){
let str='';
for(let i=0;i<fps.length;i++){
j=0;
if(i>0&&fps[i].charAt(0)=='-'){
j=1;
}
str=str+`<p>$$\\i{`+fps[i].substring(j)+`}=\\p{`+fss[i].substring(j)+`}\\C$$</p>`
}
return str;
}
function iterms1(fps,fss){
let str=itermsprep(fps,fss);
str=str+'<p>$$\\i{'+itermscombine(fps,false)+'}='+itermscombine(fss,true)+'\\C$$</p>'
return str;
}
function itermscombine(fss,ps){
if(fss.length==0){
return '';
}
let strs=['\\p{','}'];
if(!ps){
strs=['',''];
}
let str=strs[0]+fss[0]+strs[1];
for(let i=1;i<fss.length;i++){
j=0;
let s='+';
if(fss[i].charAt(0)=='-'){
j=1;
s='-'
}
str=str+strs[0]+s+strs[1]+strs[0]+fss[i].substring(j)+strs[1];
}
return str;
}
function defint1(a,b,fps,str){
let its=inttermsall(turnblack(fps));
let fs=turnblack(its[its.length-2]);
let fb='\\p{\\left[\\b{'+fs.replaceAll('x','(\\r{'+b+'})')+'}\\right]}';
let fa='\\p{\\left[\\b{'+fs.replaceAll('x','(\\pu{'+a+'})')+'}\\right]}';
let strs=['',''];
if(str.length>0){
strs=['\\t{'+str+'}&=','\\\\[10pt]'];
fps='\\g{'+fps+'}'
}
//`+strs[0]+`\\p{\\ie{\\p{\\pu{`+a+`}}}{\\p{\\r{`+b+`}}}{\\p{`+fps+`}}}`+strs[1]+`&\\p{=}\\p{\\b{`+fs+`}}\\p{\\ev{\\p{\\pu{`+a+`}}}{\\p{\\r{`+b+`}}}}\\\\[10pt]&\\p{=}`+fb+`\\p{-}`+fa+`
return `<p>$$\\a{`+strs[0]+`\\p{\\ie{\\p{\\pu{`+a+`}}}{\\p{\\r{`+b+`}}}{\\p{`+fps+`}}}`+strs[1]+`&\\p{=}\\p{\\b{`+fs+`}}\\p{\\ev{\\p{\\pu{`+a+`}}}{\\p{\\r{`+b+`}}}}\\\\[10pt]&\\p{=}`+fb+`\\p{-}`+fa+`}$$</p>`;
}
function turnblack(str){
let strs=['\\b','\\r','\\g']
for(let i=0;i<strs.length;i++){
str=str.replaceAll(strs[i],'')
}
return str;
}
function parsec(c){
if(c=='-'){
return -1;
}
return c==''?1:parseInt(c);
}
function findChar(str,chars){
let is=[];
for(let i=0;i<str.length;i++){
for(let j=0;j<chars.length;j++){
if(str[i]==chars[j]){
is.push(i)
}
}
}
return is;
}
function inttermsall(str){
str=turnblack(str)
if(str.startsWith('{')){
str=str.substring(1,str.length-1);
}
let fss='';
let wstrs='';
let ints='';
let is=findChar(str,['+','-']);
let terms=str.split(/[+-]/)
for(let i=0;i<terms.length;i++){
if(terms[i].length>0){
let strs1=intterm(terms[i])
let str1='\\p{'+strs1[0]+'}';
let str3=strs1.length>1?strs1[1]:terms[i];
fss=fss+(i>0?str[is[i-1]]:'')+str3;
//Remove extra work because studying for test now
//wstrs=wstrs+'<p>$$\\i{'+str3+'}='+str1+'$$</p>'
let str2=i>0?'\\p{'+str[is[i-1]]+'}':''
ints=ints+str2+str1;
}
}
return [wstrs,ints,fss]
}
function intterm(str){
if(str.length==0){
return '';
}
if(!str.includes('x')){
return intc(str)
}
else if(!str.includes('^')){
if(str.includes('frac')){
return intcox(str)
}
return intcx(str)
}
else if(str.includes('x^')){
return intcxn(str)
}
else if(str.includes('e^{')){
return intceax(str)
}
else{
return intetx(str)
}
}
function intc(str){
return ['\\b{'+str+'}\\p{x}','\\b{'+str+'}']
}
function intcx(str){
return intcxn(str+'^1')
}
function intcox(str){
let j=str.search('{');
let k=str.search('}')
let c=str.slice(j+1,k)
return c.length>0?['\\b{'+c+'}\\ln(x)','\\frac{\\b{'+c+'}}{x}']:['\\ln(x)','\\frac{1}{x}']
}
function intcxn(str){
let i=str.search('x\\^');
let c='\\b{'+str.slice(0,i)+'}';
if(i==0){
c='1';
}
let n='\\r{'+str.substring(i+2)+'}';
return [`\\p{\\frac{\\p{`+c+`}}{\\p{`+n+`}\\p{+}\\p{1}}}\\p{x}^{\\p{`+n+`}\\p{+}\\p{1}}`,c+'x^'+n]
}
function intceax(str){
let i=str.search('e\\^{');
let c='\\b{'+str.slice(0,i)+'}';
if(i==0){
c='1';
}
let exp=str.substring(i+3);
let j=exp.search('x')
let a='\\r{'+exp.slice(0,j)+'}';
return [`\\frac{\\p{`+c+`}}{\\p{`+a+`}}\\p{\\cdot}\\p{e^{\\p{`+a+`}\\p{x}}}`,``+c+`e^{`+a+`x}`]
}
function inttermsex(fs){
let strs=inttermsall(fs);
return {
question:'<p>Let $f(x)='+fs+`.$</p><p>Find $\\i{f(x)}$</p>`,
steps:[
strs[0]+'<p>$$\\i{'+fs+'}=\\gb{\\p{'+strs[1]+'}\\C}$$</p>'//fs instead of strs[2] because of lack of color coding and x^1
]
}
}
function inteq(fs){
return turnblack(inttermsall(fs)[1])
}
function intetx(str){
let i=str.search('e^x')
let c=parsec(str.slice(0,i-2))
let cs='\\p{\\b{'+c+'}}';
if(c==1){
cs='';
}
return [cs+'\\p{\\p{e}\\p{^x}}',cs+'e^x']
}
function areapw(a,b,c,gs,hs){
return '<p>$$\\t{Area}='+defintstr('\\pk{'+a+'}','\\r{'+c+'}','\\b{'+gs+'}')+'\\p{+}'+defintstr('\\r{'+c+'}','\\go{'+b+'}','\\g{'+hs+'}')+'$$</p>'
}
function pw(c,gs,hs){
return `$$f(x)=\\begin{cases}\\b{`+gs+`}&x<\\r{`+c+`}\\\\[4pt]\\g{`+hs+`}&x\\ge\\r{`+c+`}\\end{cases}$$`
}
function pwq(a,b,c,gs,hs,C){
return (C?'t':'Find t')+'he area under '+pw(c,gs,hs)+'from $x=\\pk{'+a+'}$ to $x=\\go{'+b+'}$'+(C?' is ':'.')
}
function pwwu(a,b,c,gs,hs,C){
return pwq(a,b,c,gs,hs,true)+areapw(a,b,c,gs,hs)
}
function defintbar(a,b,fs){
return inteq(fs)+'\\p{\\ev{\\p{'+a+'}}{\\p{'+b+'}}}'
}
function defintstr(a,b,fs){
return '\\p{\\ie{\\p{'+a+'}}{\\p{'+b+'}}{\\p{'+fs+'}}}'
}
function twodefint(a,b,c,d,gs,hs,s,str){
let strs10=str.length>0?['\\t{'+str+'}&=','\\\\[10pt]']:['',''];
s='\\p{'+s+'}';
let ints=defintstr(a,b,gs)+s+defintstr(c,d,hs);
let nextstep='&\\p{=}\\p{\\left('+defintbar(a,b,gs)+'\\right)}'+s+'\\p{\\left('+defintbar(c,d,hs)+'\\right)}\\\\[10pt]&\\p{=}'
let ans='\\p{\\left('+defintplugin(a,b,gs)+'\\right)}'+s+'\\p{\\left('+defintplugin(c,d,hs)+'\\right)}'
return `$$\\a{`+strs10[0]+ints+strs10[1]+nextstep+ans+`}$$`
}
function pwcalc(a,b,c,gs,hs){
return twodefint('\\pk{'+a+'}','\\r{'+c+'}','\\r{'+c+'}','\\go{'+b+'}','\\b{'+gs+'}','\\g{'+hs+'}','+','Area')
}
function defintplugin(a,b,fs){
let intf=inteq(fs);
return '\\p{\\left[\\p{'+intf.replaceAll('x','{('+b+')}')+'}\\right]}\\p{-}\\p{\\left[\\p{'+intf.replaceAll('x','{('+a+')}')+'}\\right]}'
}
function pwex(a,b,c,gs,hs){
return{
question:pwq(a,b,c,gs,hs,false),
steps:[
picsize('pwarea4',500)+areapw(a,b,c,gs,hs),
pwcalc(a,b,c,gs,hs)
]
}
}
function areatc(a,b,c,fs,gs,f,g){
let fc=f(c)
let gc=g(c)
a='\\go{'+a+'}'
b='\\b{'+b+'}'
let bt=fc>gc?['f(x)','g(x)',fs,gs]:['g(x)','f(x)',gs,fs]
return {
question:'Find the area between the curves $f(x)='+fs+'$ and $g(x)='+gs+'$ from $x=\\go{'+a+'}$ to $x=\\b{'+b+'}$',
steps:[
`$$\\gb{\\a{\\p{f(`+c+`)}&\\p{{}=`+eval1(fs,c)+`}\\\\[10pt]&\\p{=\\p{`+fc+`}}\\\\[10pt]\\p{g(`+c+`)}&\\p{=}\\p{`+eval1(gs,c)+`}\\\\[10pt]&\\p{=}\\p{`+gc+`}}}$$</p><p>$\\bb{\\ga{\\t{Since }\\p{`+bt[0]+`}\\p{\\t{ has the higher $y$-value, }}\\\\[4pt]\\p{\\t{the $\\gt{top}$ function is }}\\p{`+bt[0]+`=}\\p{\\g{`+bt[2]+'}}\\p{\\t{ and}}\\\\[4pt]\\p{\\t{the $\\rt{bottom}$ function is }}\\p{'+bt[1]+'=}\\p{\\r{'+bt[3]+'}}}}$',
`$$\\a{\\gb{\\t{Area}}&=\\bb{\\iep{`+a+`}{`+b+`}{\\gt{top function}}\\p{-}\\iep{`+a+`}{`+b+`}{\\rt{bottom function}}}\\\\[10pt]&=\\gb{\\iep{`+a+`}{`+b+`}{\\g{`+bt[2]+`}}\\p{-}\\iep{`+a+`}{`+b+`}{\\r{`+bt[3]+`}}}}$$`,
twodefint(a,b,a,b,'\\g{'+bt[2]+'}','\\r{'+bt[3]+'}','-','Area')
]
}
}
function limx(x,f,a){
return limxs(x,f,a,'')
}
window.j = {
startCollapsed: false,
lessonNum: 13,
lessonName: "Practice Final",
intro:String.raw`<p>The exam will have 12 questions<ol><pli>Find the limit of a piecewise function (section 1)</pli><pli>Calculate $f'(a)$ (section 2)</pli><pli>Increasing and Decreasing Intervals of a Cubic Function (section 3)</pli><pli>Fence by a River (section 4)</pli><pli>Derivatives involving $e^x$, $e^{ax}$, and $\ln(x)$ (section 5)</pli><pli>Solve for current amount (sections 6-7)</pli><pli>Solve for starting amount (section 8)</pli><pli>Area Under a Split Domain Function (section 9)</pli><pli>Area Between Two Curves With Given Endpoints (section 10)</pli><pli>Consumer Surplus (section 11)</pli><pli>Producer Surplus (section 12)</pli><pli>Volume (section 13)</pli></ol></p>`,
sections:[
{
name: String.raw`Calculating Limits of Piecewise functions`,
intro: String.raw `<p>In this section,\p we first explain\p what a piecewise function is.</p><p>Then\p we explain how to\p calculate\p limits of\p piecewise functions.</p><p><b>What is a Piecewise Function?</b></p><p>A piecewise function\p is\p an equation like this:$$`+pwfn('f','g(x)','h(x)','a','<','\\ge')+'$$'+
//Note: instead of $\r{<}$ and $\g{\ge}$ it could also be $\r{\le}$ and $\g{>}$.</p><p>The $\r{<}$ and $\g{\ge}$ could also be switched, with the $\r{<}$ on the bottom row and $\g{\ge}$ on the top row.</p><p>
`<p>This is an example of piecewise function:$$`+pwfn('f','2x+4','x^3','2','\\ge','<')+String.raw`$$\p When\p substituting an\p <span class='blue'>$\b{x}$-value</span> which is\p <span class='green'>greater than or equal to</span>\p $\go{2}$\p into\p $f(\b{x})$,\p we use\p the function\p $\g{2\b{x}+4}.$</p><p>For example,\p since \p$\p{\b{4}}\p{\g{\ge}}\p{\go{2},}$\p $f(\b{4})\p{=}\g{\p{2}\p{(\b{4})}\p{+}\p{4}}\p{.}$</p><p>When\p substituting an\p <span class='blue'>$\b{x}$-value</span>\p which \pis \p<span class='red'>less than</span> \p$\go{2}$,\p into $f(x)$\p we use\p the function\p $\r{\b{x}^3}.$</p><p>For example,\p since\p $\p{\b{1}}\p{\r{<}}\p{\go{2},}$ $\p{f(\b{1})}\p{=}\p{\r{(\b{1})^\p{3}}}\p{.}$</p>`,
//`<p><b>How to Calculate Limits of a Piecewise Function?</b></p><p>Since $\b{x}\to\go{a}^{\r{-}}$ means that $\b{x}$ is approaching $\go{a}$ and <span class='red'>less than</span> $\go{a}$, $\displaystyle\lim_{\b{x}\to \go{a}^{\r{-}}}f(x)=\lim_{\b{x}\to\go{a}^{\r{-}}}\r{g(\b{x})}$</p><p>Since $\b{x}\to\go{a}^{\g{+}}$ means that $\b{x}$ is approaching $\go{a}$ and <span class='green'>greater than</span> $\go{a}$, $\displaystyle\lim_{\b{x}\to \go{a}^{\g{+}}}f(x)=\lim_{\b{x}\to\go{a}^{\g{+}}}\g{h(\b{x})}$</p><p> To determine $\displaystyle\lim_{\b{x}\to\go{a}}f(x)$, compare $\displaystyle\lim_{\b{x}\to \go{a}^{\r{-}}}f(x)$ and $\displaystyle\lim_{\b{x}\to \go{a}^{\g{+}}}f(x)$: <ol><li>If $\displaystyle\lim_{\b{x}\to \go{a}^{\r{-}}}f(x)$ and $\displaystyle\lim_{\b{x}\to \go{a}^{\r{+}}}f(x)$ are both equal to <span class='purple'>the same number</span>, then $\displaystyle \lim_{\b{x}\to\go{a}}f(x)=\yut{that number}$</li><li>$\displaystyle\lim_{\b{x}\to \go{a}^{\r{-}}}f(x)$ and $\displaystyle\lim_{\b{x}\to \go{a}^{\r{+}}}f(x)$ are equal to different numbers, then $\displaystyle\lim_{\b{x}\to \go{a}}f(x)=\t{DNE}$</li></ol></p>`,
rightColWidth: 90,
steps: {
general: {
question: String.raw`$$`+rpp(pwfn('f','g(x)','h(x)','a','<','\\ge'))+'\\p{\\text{ OR }}'+rpp(pwfn('','h(x)','g(x)','a','\\ge','<'))+String.raw`$$\p Find each of the following:<ol><pli>$`+rp(liml('\\p{f(\\b{x})}','a'))+`$</pli><pli>$`+rp(limr('\\p{f(\\b{x})}','a'))+`$</pli><pli>$`+rp(lim('\\p{f(\\b{x})}','a'))+`$</pli><ol>`,
steps: [
`Since $\\b{x}\\to\\go{a}^{\\r{-}}$\\p means that\\p $\\b{x}$ is approaching $\\go{a}$\\p and\\p <span class='red'>less than</span>\\p $\\go{a}$, $`+liml('\\p{f(\\b{x})}','a')+`\\p{=}`+liml('\\p{\\r{g(\\b{x})}}','a')+`$</p><p>Calculate $`+liml('\\p{\\r{g(\\b{x})}}','a')+`$\\p by\\p plugging in\\p $\\got{what $x$ is approaching}$\\p into\\p $\\r{g(\\b{x})}$</p>`,
`Since $\\b{x}\\to\\go{a}^{\\g{+}}$\\p means that\\p $\\b{x}$ is approaching $\\go{a}$\\p and\\p <span class='green'>greater than</span>\\p $\\go{a}$, $`+limr('\\p{f(\\b{x})}','a')+`\\p{=}`+limr('\\p{\\g{h(\\b{x})}}','a')+`$</p><p>Calculate $`+limr('\\p{\\g{h(\\b{x})}}','a')+`$\\p by\\p plugging in\\p $\\got{what $x$ is approaching}$\\p into\\p $\\g{h(\\b{x})}$</p>`,
// String.raw`<p>Since $\lb{x}\to\go{a}^{\g{+}}$ means that $\lb{x}$ is approaching $\go{a}$ and <span class='green'>greater than</span> $\go{a}$, $$\lim_{\lb{x}\to \go{a}^{\g{+}}}f(x)=\lim_{\lb{x}\to\go{a}^{\g{+}}}\g{h(\lb{x})}$$Calculate $\lim\limits_{\b{x}\to\go{a}^{\g{+} }}\g{h(\b{x})}$ by substituting $\go{a}$ for $\b{x}$</p>`,
String.raw`<p>To determine $`+lim('\\p{f(\\b{x})}','a')+`$\\p, compare $`+liml('\\p{f(\\b{x})}','a')+`$\\p and $`+limr('\\p{f(\\b{x})}','a')+`$\\p: <ol><pli>If\\p $`+liml('\\p{f(\\b{x})}','a')+`$\\p and $`+limr('\\p{f(\\b{x})}','a')+`$\\p are\\p both\\p equal to\\p <span class='purple'>the same number</span>,\\p then $`+lim('\\p{f(\\b{x})}','a')+`\\p{=}\\p{\\yut{that number}}$</pli><pli>If\\p $`+liml('\\p{f(\\b{x})}','a')+`$\\p and $`+limr('\\p{f(\\b{x})}','a')+`$\\p are\\p equal to\\p different numbers,\\p then $`+lim('\\p{f(\\b{x})}','a')+`\\p{=}\\p{\\t{DNE}}$</pli></ol></p>`
]
},
specific: pwfnex('9+x','x^2',2,'>','\\le',11,4)
},
examples: [
pwfnex('4x+1','5',1,'<','\\ge',5,5),
pwfnex('8x','4x^2',2,'\\ge','<',16,16),
pwfnex('x^2','x',-1,'>','\\le',1,-1),
pwfnex('x-2','5x^2-2',0,'\\le','>',-2,-2),
pwfnex('x^2','5x',3,'<','\\ge',9,15),
pwfnex('x','3',-2,'<','\\ge',-2,3),
],
},
{
name: String.raw`Calculating $f'(a)$`,
intro: String.raw`<p>In this section,\p we discuss how to\p calculate\p $f'(\r{a})$</p>`,
rightColWidth: 90,
steps: {
general:{
question:String.raw`<p>Let $f(x)=\t{\{an equation\}}$</p><p>Calculate $f'(\r{a})$</p>`,
steps:[
String.raw`Calculate\p $f'(x)$`,
String.raw`Calculate\p $f'(\r{a})$\p by\p replacing\p every\p $x$\p with \p$\r{a}$`
]
},
specific:fpaex('9x^3','1')
},
examples: [
fpaex('4x^2-1','1'),
fpaex('8x+4','2'),
fpaex('x^4-4x','0'),
fpaex('3x^3-5x','-1'),
fpaex('9x^2+2x','-2'),
fpaex('8x^4+9x','4'),
],
},
{
name: String.raw`Increasing and Decreasing Graphs`,
backgroundColor: "green",
web: "Hw 3: 1, 5-7, 9; Hw 4: 1-7, 11; Hw 5: 2, 5-7; PT 2: 1-6, 8; Hw 6: 4, 7, 9-10; PT 3: 8; Hw 9: 8, 10; PF: 4, 6, 10-12, 14-15",
book: "148",
exam: "Up to one question on test 2 and up to one question on the final",
intro: String.raw`<p>`+picsize('incdecslopes',500)+String.raw`</p><p>As seen in the above picture,<ol><pli>$f(x)$ is <span class='green'>increasing</span> when<span class="invisible"> its slope is </span><span class="invisible"><span class='green'>positive</span></span></pli><pli>$f(x)$ is<span class="invisible"><span class='red'> decreasing </span>when</span><span class="invisible"> its slope is </span><span class="invisible"><span class='red'>negative</span></span></pli></ol></p><p><span class="hi">The calculus word for<span class="invisible"> slope at a point is</span><span class="invisible"> the derivative</span></span></p><p>Therefore,<ol><pli><span class="hi">$f(x)$ is $\gt{increasing}$ when <span class="invisible">$f'(x)$</span><span class="invisible"> is</span><span class="invisible"> $\gt{positive}$</span></span></pli><pli><span class="hi">$f(x)$ is $\rt{decreasing}$ when <span class="invisible">$f'(x)$</span><span class="invisible"> is</span><span class="invisible"> $\rt{negative}$</span></span></pli></ol></p>`+picsize('incdecv2',500)+`<p><b>How to find where $f(x)$ is increasing and decreasing?</b></p><p>The points where\\p $f(x)$ changes from\\p $\\gt{increasing}$\\p to\\p $\\rt{decreasing}$\\p must satisfy\\p $f'(x)=0$</p><p>We call the points where\\p $f'(x)=0$\\p $\\ot{critical points}$</p>`,
rightColWidth: 50,
steps:{
general:{question:String.raw`<p>$$f(x)=\{\t{an equation}\}$$Find the intervals where $f(x)$ is increasing and where $f(x)$ is decreasing</p>`,
steps:[
String.raw`<p>Find\p $f'(x)$</p>`,
String.raw`<p>\pSolve\p $f'(x)=0$\p to get\p the <span class='orange'>critical points</span></p>`,
String.raw`<p>Make a\p number line\p with\p the <span class='orange'>critical points</span> on it</p>`,
String.raw`<p>The critical points\p divide the number\p into\p smaller parts\p called\p intervals.</p><p>Add an\p $x$-value\p (called a \p<span class='purple'>test point</span>)\p to\p each interval</p>`,
String.raw`<p>Calculate\p $f'\p{(\put{each test point})}$\p and\p determine\p whether it is \p<span class='green'>positive</span>\p or\p <span class='red'>negative</span>.</p>`,//It is often easier to calculate using the factored version of $f'(x)$</p>`,
String.raw`<p>Add a\p $\g{+}$ sign\p above\p each interval containing\p a test point\p with a\p <span class='green'>positive</span> $f'$ value\p and a\p $\r{-}$ sign\p above \peach interval containing\p a test point\p with a \p<span class='red'>negative</span> $f'$ value</p>`,
String.raw`<p>Determine the intervals where\p $f'(x)$\p is\p <span class='green'>positive</span>\p and\p <span class='red'>negative</span><ol><pli>$f'(x)\g{>0}$ in<span class="invisible"> the intervals</span><span class="invisible"> with a</span><span class="invisible"> $\g{+}$ sign </span><span class="invisible">above them</span></pli><pli>$f'(x)\r{<0}$ in<span class="invisible"> the intervals</span><span class="invisible"> with a</span><span class="invisible"> $\r{-}$ sign</span><span class="invisible"> above them.</span></pli></ol></p><p><b>Note:\p The endpoints of\p the intervals are the\p $\btip{\o{\textbf{critical points}}}{x\t{-values where }f'(x)=0}$ \pnot\p the $\btip{\pu{\textbf{test points}}}{\t{the }x\t{-values on each side of the critical points}}$</b></p>`,
String.raw`<p>Determine\p the intervals where\p $f(x)$ is\p <span class='green'>increasing</span>\p and\p <span class='red'>decreasing</span><ol><pli>$f(x)$ is <span class="invisible"><span class='green'>increasing</span></span><span class="invisible"> on </span><span class="invisible">the intervals where</span><span class="invisible"> $f'(x)\g{>0}$</span></pli><pli>$f(x)$ is <span class="invisible"><span class='red'>decreasing</span></span><span class="invisible"> on </span><span class="invisible">the intervals where</span><span class="invisible"> $f'(x)\r{<0}$</span></pli></ol></p><p><b>Note:\p The endpoints of\p the intervals are the\p $\btip{\o{\textbf{critical points}}}{x\t{-values where }f'(x)=0}$\p not\p the $\btip{\pu{\textbf{test points}}}{\t{the }x\t{-values on each side of the critical points}}$</b></p>`,
]
},
specific:incdecsetup(1,-1,1,0)
//incdec(`2x^3+6x^2+6x`,`6x^2+12x+6`,x=>6*Math.pow(x,2)+12*x+6,String.raw`6(x^2+2x+1)&=0\\[4pt]6(x+1)(x+1)&=0\\[4pt]x+1=0,\ x+1&=0`,[-1],[-2,0]),
},
examples: [
incdecsetup(1,-2,2,0),
incdecsetup(-1,-2,2,1),
incdecsetup(1,-4,0,0),
incdecsetup(-1,-1,3,1),
incdecsetup(-2,-1,1,5),
incdecsetup(-1,-3,1,-4)
],
},
{
name: String.raw`A Fence by a River`,
intro:String.raw`<p>In this section,\p we will describe how to\p answer the following question:</p><p>A farmer wants to build\p a three-sided\p rectangular fence\p near a river,\p using $\yk{d}$ yards of fencing.</p><p>Assume that\p the river runs straight\p and that\p John need not fence in\p the side next to the river.</p><p>What are the\p $\yut{length}$\p and\p the $\ot{width}$ of\p the fence which\p maximizes the enclosed area?</p><p>A diagram for the problem,\p provided on the exam but not in WebAssign,\p is below:
</p><p><iframe src="https://www.desmos.com/calculator/hbi9py1k7h?embed" width="500" height="500" style="border: 1px solid #ccc" frameborder=0></iframe></p><p>
We solve this problem\p in two stages:</p><p><ol><pli>Rewrite <span class="invisible">the word problem as</span><span class="invisible"> maximizing a function $f(x)$</span></pli><pli>Use the steps<span class="invisible"> from the previous lecture</span><span class="invisible"> to </span><span class="invisible">maximize $f(x)$</span></pli></ol></p><p>In the first stage,</p><p><ol><pli>Write the area in terms of<span class="invisible"> $\ot{width}$</span><span class="invisible"> and</span><span class="invisible"> $\yut{length}$</span></pli><pli>Use the<span class="invisible"> $\btip{\ykt{amount of fence used}}{\t{given in the problem}}$ </span><span class="invisible">to </span><span class="invisible">write $\yut{length}$ </span><span class="invisible">in terms of</span><span class="invisible"> $\ot{width}$.</span></pli><pli>Use the previous two steps to<span class="invisible"> rewrite area</span><span class="invisible"> in terms of</span><span class="invisible"> only $\ot{width}$,</span><span class="invisible"> i.e.,</span><span class="invisible"> $\t{area}=f(\ot{width})$</span></pli></ol></p><p> In the second stage,\p we maximize\p $f(\ot{width})$ by</p><p>
<ol>
<pli>
Finding<span class="invisible"> the critical point</span>
</pli>
<pli>
Using <span class="invisible">$f''(x)$</span><span class="invisible"> to </span><span class="invisible">verify that</span><span class="invisible"> the critical point is</span><span class="invisible"> a max</span>
</pli>
<pli>
Use <span class="invisible">the critical point</span><span class="invisible"> to</span><span class="invisible"> find the</span><span class="invisible"> $\yut{length}$</span>
</pli>
<pli>
In WebAssign,<span class="invisible"> you will have to</span><span class="invisible"> plug </span><span class="invisible">the $\ot{width}$</span><span class="invisible"> and </span><span class="invisible">$\yut{length}$</span><span class="invisible"> into</span><span class="invisible"> the area equation</span><span class="invisible"> to get</span><span class="invisible"> the maximum area.</span>
</pli>
</ol></p>`,
rightColWidth: 90,
steps: {
general:{
question:fenceQ('d'),
steps:[
String.raw`Write\p an equation for\p the area \pin terms of\p $\ot{width}$\p and\p $\yut{length}$`,
String.raw`Use\p the $\pkt{amount of fence used}$\p to write\p an equation that\p relates\p $\ot{width}$\p and\p $\yut{length}$`,
String.raw`Solve\p the equation from step 2\p for $\yut{length}$\p in terms of\p $\ot{width}$</p><p>It is equally correct to\p solve for\p $\ot{width}$\p in terms of\p $\yut{length}$\p if you prefer that`,
String.raw`Substitute\p the equation from step 3\p for $\yut{length}$\p into\p the area equation`,
String.raw`Find $f'(x)$</p><p>I recommend\p distributing the\p $x$\p to avoid product rule<br>Using product rule is\p totally correct\p but it takes longer`,
String.raw`Solve\p $f'(x)=0$\p to get\p the $\btip{\ot{critical point}}{\b{x\t{-value where }f'(x)=0}}$`,
String.raw`Find $f''(x)$`,
String.raw`<p>Determine\p if $f''(x)$ is\p always $\gt{positive}$\p or\p always $\rt{negative}$</p><p><ol><pli>If all coefficients<span class="invisible"> in $f''(x)$</span><span class="invisible"> are $\gt{positive}$,</span><span class="invisible"> $f(x)$ is</span><span class="invisible"> concave $\gt{up},$</span></span><span class="invisible"> so $f(x)$ has</span><span class="invisible"> a $\gt{min}$ at</span><span class="invisible"> and</span><span class="invisible"> no $\rt{max}$</span></pli><pli>If all coefficients<span class="invisible"> in $f''(x)$</span><span class="invisible"> are $\rt{negative}$,</span><span class="invisible"> $f(x)$ is </span><span class="invisible"> concave $\rt{down},$</span><span class="invisible"> so $f(x)$ has</span><span class="invisible"> a $\rt{max}$</span><span class="invisible"> and</span><span class="invisible"> no $\gt{min}$</span></pli></ol></p>`,
String.raw`Find the\p $\yut{length}$ of\p the fence by\p plugging in\p the $\ot{critical point}$\p for $x$\p into\p the equation from step 4`,
// String.raw`Find\p the area by multiplying\p $\ot{width}$\p by\p the $\put{length}$`
]
},
specific:fence(100)
},
examples: [
fence(16),
fence(8),
fence(12),
fence(20),
fence(24),
fence(20),
//T
// fence(40),
// fence(20)
],
},
{
name: String.raw`Derivatives involving $e^x$, $e^{ax}$ and $\ln(x)$`,
backgroundColor: "green",
web: "Hw 3: 1, 5-7, 9; Hw 4: 1-7, 11; Hw 5: 2, 5-7; PT 2: 1-6, 8; Hw 6: 4, 7, 9-10; PT 3: 8; Hw 9: 8, 10; PF: 4, 6, 10-12, 14-15",
book: "148",
exam: "Up to one question on test 2 and up to one question on the final",
intro: String.raw`<p>In this section,\p we will discuss how to\p take derivatives involving\p $e^x$, \p $e^{ax}$,\p and\p $\ln(x)$</p><p>We will use the following derivative rules</p><p><table>
<tr>
<th>Formula</th>