Question 1019976
{{{f(x)=(x^2-1)/(x-3)}}}
 
a. Since the expression for {{{(x^2-1)/(x-3)}}} is defined for all real values of {{{x}}} such that {{{x<>3}}} ,
the domain for the function in interval notation is
{{{"("}}}{{{-infinity}}}{{{", 3 ) U ( 3 ,"}}}{{{infinity}}}{{{")"}}} .
 
b. Find all vertical, horizontal and slant asymptotes
{{{x=3}}} is a vertical asymptote,
because the function does not exist for {{{x=3}}} ,
but {{{lim(x->3,abs((x^2-1)/(x-3)))=lim(x->3,abs((3^2-1)/(x-3)))=lim(x->3,abs(8/(x-3)))=infinity}}} .
{{{f(x)=(x^2-1)/(x-3)=(x^2-9+8)/(x-3)=((x+3)(x-3)+8)/(x-3)=(x+3)(x-3)/(x-3)+8/(x-3)=x+3+8/(x-3)}}} ,
so {{{highlight(y=x+3)}}} is the slant asymptote, because
{{{lim(x->infinity,f(x)-(x+3))=lim(x->infinity,8/(x-3))=0}}} .
That means the graph function "hugs" the slanted line representing {{{y=x+3}}} towards the left and right ends of the graph, and off course, there is no horizontal asymptote.
We even know that {{{f(x)-(x+3)=8/(x-3)>0}}} for {{{x>3}}} , meaning the curve is above the asymptote,
and {{{f(x)-(x+3)=8/(x-3)<0}}} for {{{x<3}}} , meaning the curve is below the asymptote.
 
c. Sketch the function
We know the asymptotes.
To sketch it, we may want to know
1)where the function is negative, zero, and positive, and
2) maxima and minima.
Since {{{f(x)=(x^2-1)/(x-3)=(x+1)(x-1)/(x-3)}}} ,
{{{x=-1}}} and {{{x=1))) make the numerator, and {{{f(x)}}} zero.
For {{{x<-1}}} , {{{system(x+1<0,x-1<0,x-3<0)}}} ---> {{{f(x)<0}}} .
For {{{-1<x<1}}} , {{{system(x+1>0,x-1<0,x-3<0)}}} ---> {{{f(x)>0}}} .
For {{{1<x<3}}} , {{{system(x+1>0,x-1>0,x-3<0)}}} ---> {{{f(x)<0}}} .
For {{{x>3}}} , {{{system(x+1>0,x-1>0,x-3>0)}}} ---> {{{f(x)>0}}} .
To find maxima and minima, we need to calculate the derivative:
{{{f(x)=x+3+8/(x-3)}}} , so {{{"f ' ( x )"=1-8/(x-3)^2=((x-3)^2-8)/(x-3)^2=(x^2-6x+1)/(x-3)^2}}} .
{{{"f ' ( x )"=0}}}--->{{{(x-3)^2=8}}}--->{{{x=3 +- sqrt(8)=x=3 +- 2sqrt(2)}}} .
Approximate values are {{{3-sqrt(8)=0.17}}} and {{{3+sqrt(8)=5.83}}} .
So {{{f(3-sqrt(8))=about(0.17^2-1)/(0.17-3)=about0.34}}} and
{{{f(3+sqrt(8))=about(5.83^2-1)/(5.83-3)=about11.66}}} .
We can graph the asymptotes, extreme value points, and zeros:
{{{drawing(300,450,-10,10,-7,23,grid(0),
green(line(-10,-7,10,13)),green(circle(-1,0,0.2)),
green(circle(1,0,0.2)),green(circle(0.17,0.34,0.2)),
green(circle(5.83,11.66,0.2)),green(line(3,-10,3,25))
)}}} ,  and with all else we found about {{{f(x)}}} we sketch like this: {{{drawing(300,450,-10,10,-7,23,graph(300,450,-10,10,-7,23,(x^2-1)/(x-3)),
green(line(-10,-7,10,13)),green(circle(-1,0,0.2)),
green(circle(1,0,0.2)),green(circle(0.17,0.34,0.2)),
green(circle(5.83,11.66,0.2)),green(line(3,-10,3,25))
)}}}
 
d. Solve the inequality 
{{{(x^2-1)/(x-3)<1}}}
We could look at the graph and see that the solution is {{{x<3}}} .
Without the graph,
{{{(x^2-1)/(x-3)<1}}}<-->{{{(x^2-1)/(x-3)-1<0}}}<-->{{{(x^2-1)/(x-3)-(x-3)/(x-3)<0}}}<-->{{{(x^2-x+2)/(x-3)<0}}}
Since the denominator is always positive, the solution is {{{x-3<0}}}<-->{{{x<3}}} .