Maple Support For Exam III

Problems 10, 7, 4, 5

>	with(plots):

Warning, the name changecoords has been redefined

Number 10

>	f:=(x^4+x^3-7*x^2-x+6)/(x^2-9);

>	factor(x^4+x^3-7*x^2-x+6);

>	(x-1)*(x-1)*(x+3)*(x+1);

>	factor(x^2-9);

>	simplify(((x-1)*(x-2)*(x+3)*(x+1))/((x-3)*(x+3)));

Note that x cannot equal -3 in the simplification above.

>	fprime:=diff(f,x);

>	Q:=quo(x^4+x^3-7*x^2-x+6,x^2-9,x);

>	fGraph:=plot(f,x=-10..10,y=-100..100,thickness=2,discont=true):

>	Qgraph:=plot(x^2+x+2,x=-10..10,y=-100..100,thickness=2,color=blue):

>	Limit(f,x=-3);

>

value(%);

We have a removable discontinuity at (-3,20/3).

>	RemovDiscont:=pointplot([-3,20/3],symbol=circle,symbolsize=16,color=green):

>	VertAsym:=implicitplot(x=3,x=-10..10,y=-100..100,thickness=2,color=blue):

The function is graphed below in red along with the vertical asymptote in blue, the "asymptotic parabola" in blue, and the removable discontinuity in green.

>	display(fGraph,Qgraph,VertAsym,RemovDiscont);

>	display(fGraph);

Notice that if we do not set "discont=true" Maple attempts to connect points on each side of a vertical asymptote.  The result is to draw nearly vertical lines that appear to be part

of the graph of the function.

>	fGraph2:=plot(f,x=-10..10,y=-100..100,thickness=2):

>	display(fGraph2);

Let's look at a "blow-up" of the graph around some of the critical points.

>	plot(f,x=-3..3,y=-3..3,thickness=2);

>	fsolve({fprime},{x},-1..1);

>	eval(f,x=-.7771830483e-1);

>	fsolve({fprime},{x},1..2);

>	eval(f,x=1.629443541);

>	fsolve({fprime},{x},3..5);

>	eval(f,x=3.948274764);

We have relative minimum points at approximately (-0.0771830483,-0.6710063760) and (3.948274764,29.97352179 and a relative maximum point at approximately

(1.629443541,0.4474845938).

>	fPP:=diff(f,x,x);

>	fsolve({fPP},{x},0..2);

>	eval(f,x=1);

>	eval(fPP,x=0.99);

>	eval(fPP,x=1.01);

We have an inflection point at (1,0).

>	eval(f,x=-3.01);

>	eval(f,x=-2.99);

>	plot(f,x=-3.01..-2.99,y=6.614540735..6.718985191,thickness=10,discont=true);

Number 7

>	f7:=(3*x^2+5*x-8)/(x+2);

>	f7Graph:=plot(f7,x=-6..6,y=-50..50,thickness=2,discont=true):

>	f7Q:=quo(3*x^2+5*x-8,x+2,x);

>	f7QGraph:=plot(f7Q,x=-6..6,y=-50..50,thickness=2,color=blue):

>	f7VertAsym:=implicitplot(x=-2,x=-6..6,y=-50..50,thickness=2,color=blue):

>	display(f7Graph,f7QGraph,f7VertAsym);

Number 4

>	f4:=x^3/(x-2);

>	f4Graph:=plot(f4,x=-6..6,y=-50..50,thickness=2,discont=true):

>	f4Q:=quo(x^3,x-2,x);

>	f4QGraph:=plot(f4Q,x=-6..6,y=-50..50,thickness=2,color=blue):

>	f4VertAsym:=implicitplot(x=2,x=-6..6,y=-50..50,thickness=2,color=blue):

>	display(f4Graph,f4QGraph,f4VertAsym);

>	f4P:=diff(f4,x);

>	solve({f4P},{x});

>	eval(f4,x=0);

>	eval(f4,x=3);

There are critical points at (0,0) and (3,27).  The critical point (3,27) is a relative minimum point.  The critical point (0,0) is not a relative minimum or a relative maximum.  It looks like it is

an inflection point.

>	f4PP:=diff(f4,x,x);

>	solve({f4PP},{x});

>	eval(f4PP,x=-1);

>	eval(f4PP,x=1);

Clearly the point (0,0) is an inflection point.

Number 5

>	f5:=x^3-4*x^2+4*x;

>	f5Graph:=plot(f5,x=-2..4,y=-10..10,thickness=2):

>	display(f5Graph);

>	f5P:=diff(f5,x);

>	solve({f5P},{x});

>	eval(f5,x=2/3);

>	eval(f5,x=2);

The critical point (2/3,32/27) is a relative maximum point and the critical point (2,0) is a relative minimum point.

>	f5PP:=diff(f5,x,x);

>	solve({f5PP},{x});

>	eval(f5,x=4/3);

We have an inflection point at (4/3,16/27).

>	CritInfPoints:=pointplot([[0,0],[2/3,32/27],[4/3,16/27],[2,0]],symbol=circle,symbolsize=16,color=green):

>	display(f5Graph,CritInfPoints);

>