Calculus UNIT202 IVT and Continuity! -- Sage

#1) IVT: why do you get 3 zeros? f(x)=5*x^3-5*x show(f(x)) show(factor(f(x))) show(solve(f(x)==0,x)) plot(f(x))

 $\newcommand{\Bold}[1]{\mathbf{#1}}5 \, x^{3} - 5 \, x$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}5 \, {\left(x + 1\right)} {\left(x - 1\right)} x$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[x = \left(-1\right), x = 1, x = 0\right]$

#2) AP Calculus AB FRQ: 76AB2 #2a) f(x)=x^3-3*x^2-4*x+12 #2b) find the zeros f(x)=x^3-3*x^2-4*x+12 show(f(x)) show(factor(f(x))) show(solve(f(x)==0,x)) plot(f(x),-4,4)

 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{3} - 3 \, x^{2} - 4 \, x + 12$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}{\left(x + 2\right)} {\left(x - 2\right)} {\left(x - 3\right)}$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[x = \left(-2\right), x = 2, x = 3\right]$

#2) AP Calculus AB FRQ: 76AB2 #2b) h(x)=f(x)/(x-3) when x!=3 #2b) h(x)=p when x=3 #2b) find p to make h continuous h(x)=f(x)/(x-3) show(h(x)) show(factor(h(x))) show(solve(h(x)==0,x)) p1=plot(h(x),-4,3,color='red') p2=plot(h(x),3,4,color='green') show(p1+p2) #2b) h(3)=p #2b) lim(h(x),x=3,dir='left')=lim(h(x),x=3,dir='right')=5 #2b) p=5

 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{x^{3} - 3 \, x^{2} - 4 \, x + 12}{x - 3}$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}{\left(x + 2\right)} {\left(x - 2\right)}$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left[x = \left(-2\right), x = 2\right]$

#2) AP Calculus AB FRQ: 76AB2 #2c) h(x)=f(x)/(x-3) when x!=3 #2c) is h(x) an even function? show(h(x)) show(h(-x)) #yes!

 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{x^{3} - 3 \, x^{2} - 4 \, x + 12}{x - 3}$ 
 $\newcommand{\Bold}[1]{\mathbf{#1}}\frac{x^{3} + 3 \, x^{2} - 4 \, x - 12}{x + 3}$

Calculus UNIT202 IVT and Continuity!

3369 days ago by MATH4R2013