Curve Sketching -- Extrema & Curvature Etc...

Monotonicity & Extrema

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Definition : (pg. 277) f has an Absolute Maximum value at c if for all x in the domain of f . f(c) is the maximum value of f on its domain. Similar definition for absolute minimum value. These values, if they exist, are the absolute extreme values or absolute extrema of f.

e.g. has an absolute minimum value at x = 0 and no absolute maximum value.

Definition : (pg. 278) f has a local maximum ( relative maximum ) at (c,f(c)) if there is an open interval I containing c such that for all c I. Similar definition for local (relative) minimum . For example, has a local (and absolute!) minimum at (0,0).

Extreme Value Theorem : If f is continuous on the closed interval [a,b] then f attains an absolute maximum and minimum on [a,b].

Proof: Gift from God -- beyond the level of the course. Done in Analysis I at university.

Consider the realtionship between the graphs of f and f_prime for the following:

> f:=surd(x^2-1,3);f_prime:=diff(f,x);

Note: I really wanted to write f:= but for mysterious reasons Maple will only treat this as a real-valued function for . "surd( ,3)" means the third root of for all R eal numbers.

> plot({f,f_prime},x=-4..4,y=-3..3,colour=[black,red],discont=true,thickness=2);

Fermat's Theorem : (pg. 280) If f has a local extremum at c and f'(c) exists then f'(c) = 0.

Proof: For local maximum, see blackboard.

For local minimum: exercise!

Note Bene: The converse of Fermat's theorem is not true . See below.

> plot(x^3,x=-4..4,y=-8..8,colour=black);

You should quickly see that y = satisfies y' = 0 at (0,0) but there is no relative extremum at (0,0).

Definition : c is a critical number of f if c is in the domain of f and either f'(c) = 0 or f'(c) does not exist . It follows from Fermat's Theorem that if f has a local extremum at c then c is a critical number of f .

e.g. Find the critical numbers of f(x) = .

> f:=(x^2-1)^(1/3);

> fprime:=diff(f,x);

> simplify(%);

==> the three critical numbers are -1, 0, 1.

> plot({surd(x^2-1,3),diff(surd(x^2-1,3),x)},x=-4..4,y=-3..3,colour=[black,red]);

Recipe for finding the absolute extreme values of a continuous function f on the closed interval [a,b]: (pg. 282)

1) Find the critical numbers of f on (a,b).

2) Find f(a), f(b), and f(c) for each critical number c in (a,b).

3) The largest value obtained in 2) is the absolute maximum of f on [a,b], the smallest value is the absolute minimum value of f on [a,b].

e.g. Find the absolute extreme values of f(x) = on [-2,1/2].

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

> plot(f(x),x=-2..1/2,colour=blue);

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

> solve(fprime);

> [f(-2),f(-1),f(0),f(1/2)];

==> absolute maximum value is 5 (and occurs when x = -1 and x = 1/2) and absolute minimum value is 0 (and occurs when x = 0).

> plot({f(x),fprime},x=-2..2,y=-2..8,colour=[blue,red],thickness=2);

Mean Value Theorem and the First Derivative

e.g.

Consider f(x) = 1/x on [1,2] and find c (1,2) such that f'(c) = .

> f:=x->1/x;

> (f(2)-f(1))/(2-1);

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

> solve(fprime=-1/2,x);

> evalf(%,4);

So, we see that if c = then f'(c) = = and (1,2).

> line1:=y-1=(-1/2)*(x-1);line2:=y-1/sqrt(2)=(-1/2)*(x-sqrt(2));

> l1:=solve(line1,y);l2:=solve(line2,y);

> plot({f(x),l1,l2},x=1..2,y=1/2..1,colour=[blue,red],thickness=2);

Mean Value Theorem : (Page (289) If f is continuous on [a,b] and differentiable on (a,b) then there exists (at least one) a number c (a,b) such that f'( c ) = .

Proof: Depends on the extreme value theorem -- take Analysis I in university!

Rolle's Theorem : If it happens that in the M.V.T. then we get a special case: If f is continuous on [a,b] and differentiable on (a,b) and then there exists a number c (at least one) (a,b) such that f'( c ) = 0.

The next two results are very important in Integral Calculus (Cal II):

Theorem : If f'(x) = 0 for all x (a,b) then f is constant on (a,b).

Proof: Direct consequence of the MVT.

Corollary : If f'(x) = g'(x) for all x (a,b) then f(x) = g(x) + C where C is a constant.

Proof: Do it!

Definition : f is increasing on the interval I if whenever a < b, f(a) < f(b).

f is decreasing on the interval I if whenever a < b, f(a) > f(b).

A function that is either increasing or decreasing (but not both) on I is monotonic on I.

Example.

> with(plots):P1:=plot(cot(x),x=0..Pi,y=-4..4,discont=true,colour=black,thickness=2):P2:=plot(arccot(x),x,colour=red,thickness=2):P3:=plot(x,x,colour=blue):

> display(P1,P2,P3);

> h:=arctan(x)+arccot(x);Diff(h,x)=diff(h,x);

> subs(x=1,h)=simplify(subs(x=1,h));

=> We have proved the identity

Demo

Consider the graph of :

> restart:f:=x->x^3+x^2;fprime:=diff(f(x),x);

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

> solve(fprime=0);

> slope:=unapply(fprime,x);

> tline:=a->f(a)+slope(a)*(x-a);

Now, cosider the graph, together with some of its tangent lines from -2 to -2/3:

> plot([f(x),tline(-2),tline(-1.5),tline(-1),tline(-2/3),tline(-1/2)],x=-2..2,y=-4..1/2,colour=[black,red,red,red,red,red],thickness=1);

By considering slopes, what do you notice about the derivative function as x goes from -2 to -2/3??

> [slope(-2),slope(-1.5),slope(-1),slope(-2/3)];

Now consider the graph and some tangent lines from -2/3 to 0:

> plot({f(x),tline(-0.6),tline(-1/2),tline(-1/3),tline(-.1)},x=-2/3..0,colour=black);

What do you notice about the derivative on this interval?

> [slope(-.6),slope(-0.5),slope(-1/3),slope(-.1)];

These types of considerations lead us to the following test....

Test for Monotonic functions : (pg. 294) If f is continuous on [a,b] and differentiable on (a,b) then:

1) If f'(x) > 0 for all x (a,b) then f is increasing on (a,b)

2) If f'(x) < 0 for all x (a,b) then f is decreasing on (a,b) .

Proof: MVT.

e.g. Determine where is monotonic (i.e. where it is increasing and decreasing).

> f:=surd(x,5)*(x+1);fprime:=diff(f,x);

> simplify(fprime);

> solve(fprime=0);

> fsolve(fprime=0);

> plot({f,fprime},x=-2..2,y=-1..2,colour=[black,red],thickness=2);

Clearly, f is decreasing on ( ) and increasing on ( ) .

First Derivative Test : (pg. 295) Suppose c is a critical number of f, then

1) If f' changes from positive to negative at c then (c,f(c)) is a local (relative) maximum point

2) If f' changes from negative to positive at c then (c,f(c)) is a local (relative) minimum point

3) If f' does not change sign at c then there is no local extrema at (c,f(c)).

e.g. Classify the local extrema of .

> g:=2*x^2-x^4;gprime:=diff(g,x);

> solve(gprime=0);

> plot(g,x=-2..2,y=-2..2,colour=black);

> plot(gprime,x=-2..2,y=-2..2,colour=black);

e.g. Classify the local extrema of .

> h:=x^2-1/x;hprime:=diff(h,x);

> sols:=solve(hprime=0);

> evalf(%,4);

Key numbers of h': 0,

> plot(h,x=-2..2,y=-5..5,discont=true,colour=black);

> subs(x=sols[1],h);evalf(%,4);

There is a local minimum at about (-0.79,1.89).

e.g. Show that has no relative extrema.

> m:=x-1/x;mprime:=diff(m,x);

> solve(mprime=0);

> plot(m,x=-4..4,y=-4..4,discont=true,colour=black);

Note : m'(x) > 0 for all x (except x = 0) so m is increasing on its entire domain and so cannot have any local extrema.

e.g.

Sketch a function that satisfies: f(1) = 5, f(4)=2, f'(1) = 0 = f'(4), f'(x) > 0 for x < 1, f'(x) <= 0 for x > 1.

Concavity

Demo

Consider (again) the behaviour of the graph of :

> restart:f:=x->x^3+x^2;fprime:=diff(f(x),x);

> slope:=unapply(fprime,x);

> tline:=a->f(a)+slope(a)*(x-a);

> plot([f(x),tline(-2),tline(-1.5),tline(-1),tline(-1/2)],x=-2..0,y=-4..1/2,colour=[black,red,red,red,red],thickness=2);

Definition : The graph of is called concave down on I if the graph of is below all of its tangent lines on I.

By considering slopes , what do you conclude about the behaviour of the derivative of the derivative function on the interval from -2 to -1/3?

> slope(-2),slope(-1.5),slope(-1.0),slope(-.5),slope(-0.333);

> ff:=diff(fprime,x);

> solve(ff);

> fdp:=unapply(ff,x); # fdp stands for "f double prime."

> fdp(-2),fdp(-1.5),fdp(-1),fdp(-1/2),fdp(-1/3);

Now lets consider the function from -1/3 onward...

> plot([f(x),tline(-1/3),tline(0),tline(1/3),tline(1/2),tline(1)],x=-1..2,y=-2..4,colour=[blue,red,red,red,red,red],thickness=2);

Definition: The graph of is called cancave up on I if the graph of is above all of its tangent lines on I.

By considering slopes, what do you conclude about the behaviour of the derivative of the derivative function on the interval from -1/3 onward?

> slope(-1/3),slope(0),slope(1/3),slope(1/2),slope(1);

> fdp(-1/3),fdp(0),fdp(1/3),fdp(1);

Testing for Concavity

Theorem : (pg. 298) Suppose f'' exists over the interval I, then:

a) If f'' > 0 for all x I then f is concave up on I

b) If f'' < 0 for all x I then f is concave down on I .

Proof : See the blackboard . I'll do it for f'' > 0 then you will do it for f'' < 0. Oh boy!

Definition : P is an inflection point of if the graph changes from concave up to concave down or vice versa at P. i.e. if f'' changes sign at P.

e.g.

Analyze as completely as possible , the graph of by considering the first and second derivatives of h over the intervals determined by their respective key numbers . The key numbers of a function are the numbers that make the function 0 or undefined.

> h:=x->3*x^5-5*x^3+3;hp:=x->15*x^4-15*x^2;hdp:=x->60*x^3-30*x;

> solve(h(x)=0);solve(hp(x)=0);solve(hdp(x)=0);

> fsolve(h(x)=0);

> evalf(solve(hdp(x)=0),4);

==> f is increasing on ( ,-1) U (1, ) and decreasing on (-1,1)

f is concave up on ( ,0) U ( , ) and concave down on ( , ) U (0, )

(-1,5) is a relative maximum , (1,1) is a relative minimum and there are inflection points at ( , ) and at ( , ).

> plot(h(x),x=-2..2,y=-8..8,colour=black);

Second Derivative Test

Second Derivative Test : If f'' is continuous on an open interval I containing c and f'(c) = 0, then:

(a) If f''(c) > 0 then (c,f(c)) is a local minimum point

(b) If f''(c) < 0 then (c,f(c)) is a local maximum point

(c) If f''(c) = 0 then you have no information and must use the first derivative test !

Proof: by an intuitive argument! Watch.

e.g. Classify the extrema of .

> restart:f1:=x->x^(2/3)*(6-x)^(1/3);

> plot(f1(x),x,colour=black);

> f2:=x->surd(x^2,3)*surd(6-x,3);

> plot(f2(x),x,colour=black);

These two plots should convince you that it is f2(x) that is the function we need. Why??

> f2p:=diff(f2(x),x);simplify(%);

> solve(f2p=0);

> fsolve(f2p=0);

> f2dp:=diff(f2p,x);simplify(%);

=> x = 0, 6, 4 are all critical numbers but only x = 4 qualifies for the second derivative test:

> subs(x=4,f2dp);evalf(%);

> f2(4);

=> there is a local maximum at (4, )

For the other two values, you must resort to the first derivative test (i.e. make a sign table). Do it!!

The plot from earlier "says" that (0,0) is a local minimum and no other extrema.

e.g. Use Calculus to analyze the extrema of .

> restart:f:=a*x^2+b*x+c;

> fp:=diff(f,x);fdp:=diff(fp,x);

> number:=solve(fp=0,x);

> subs(x=number,fdp);

i.e. no matter what you plug in, the second derivative is always 2a. In other words, the sign of the second derivative depends only on a. Therefore:

1) If a > 0, f''(x) > 0 so there is a local minimum at ( , f( )) -- i.e. the vertex is the lowest point

2) If a < 0, f''(x) < 0 so there is a local maximum at ( , f( )) -- i.e. the vertex is the highest point

Curve Sketching Strategy

See text, section 4.5

A) Domain : Always calculate the domaim of your function to avoid the embarrassment of drawing a sketch where the function does not exist !

B) Intercepts : a) The y -intercept is always available (when it exists) and easy: just find f(0)!

b) The x-intercept(s), when they exist, are found by solving f(x) = 0. They may be difficult or impossible to find exactly. You are expected to to find them whenever "reasonable." This entails you being able to recognize the kinds of expressions and equations that are "do-able" at your level. Practice!

C) Symmetry : If you know about symmetry, feel free to exploit it but you are not expected to use this routinely.

D) Asymptotes : Always.

a) Horizontal Asymptotes : and

[Note: 1) No polynomial has H.A.'s

2) = is not always true. ]

b) Vertical Asymptotes : Look for these where the function becomes undefined . Always show two-sided limits.

(Note: no polynomial has H.A.'s. Why?)

c) Slant Asymptotes are not part of a regular course.

E & F) Intervals of Monotonicity and Extrema : Use f' and its key numbers to make a sign table to determine the intervals of monotonicity and classify the extrema all at once.

G) Concavity and Inflection Points : Use f'' and its key numbers to make a sign table to find the intervals over which f is concave up, concave down and to determine any inflection points. Note itis possible for the concavity to change from one open interval to the next without ther being an inflection point! How??!!

*** See worked out examples 1, 2, 3, 4, 5, 6 in section 4.5 of your text ***

e.g. Sketch f( x ) =

> restart:f:=(2*x^2-8)/(x^2-16);fp:=diff(f,x);fdp:=diff(fp,x);

> simplify(fp);simplify(fdp);

> solve(f);solve(fp);solve(fdp);

> Limit(f,x=infinity)=limit(f,x=infinity);Limit(f,x=-infinity)=limit(f,x=-infinity);

> Limit(f,x=-4,left)=limit(f,x=-4,left);Limit(f,x=-4,right)=limit(f,x=-4,right);

YOU do and !

> plot({f,2},x=-10..10,y=-8..8,colour=black);

e.g. Sketch

> g:=x*ln(x);gp:=diff(g,x);gdp:=diff(gp,x);

> solve(g);solve(gp);solve(gdp);

> Limit(g,x=0,right)=limit(g,x=0,right);

> plot(g,x=0..4,y=-1..4,numpoints=1000,colour=black);

> subs(x=exp(-1),g);simplify(%);

e.g. Sketch a function that satisfies the following:

f(-1) = f(2) = -1, f(-3) =4

f'(-1) = f'(2) = 0, f'(x) = 0 if x < -3

f'(x) < 0 on (-3,-1) and (0,2)

f'(x) > 0 on (-1,0) and (2, )

f''(x) > 0 on (-3,0) and (0,5)

f''(x) < 0 on (5, ).

Applied Max-Min Problems

** Read Section 4.7 and e.g.'s 1, 2, 3, 4, 5. **

The following fact can come in handy when looking for an absolute extreme value on an open interval:

If c is a critical number of the continuous function and f'(x) > 0 for all x < c and f'(x) < 0 for all x > c then f(c) is the absolute maximum value of f. Similar statement for absolute minimum value.

e.g. #7, page 335.

A farmer with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible area of the four pens.

> Area:=x->x*(750-5*x)/2;AreaP:=diff(Area(x),x);AreaDP:=diff(AreaP,x);

> plot(Area(x),x=-10..160,colour=black);

> solve(AreaP=0);

> subs(x=75,AreaDP);

> MaxArea:=Area(75);

> evalf(%,5);

The largest possible area is about 14 thousand square feet.

e.g. Find the point(s) on closest to (2,0)

> with(plots):

> implicitplot(y^2-x^2=4,x=-4..4,y=-4..4,colour=black);

> distance:=sqrt(2*x^2-4*x+8);

> plot(distance,x=-4..4,y=0..4,colour=black);

> distanceP:=diff(distance,x);distanceDP:=diff(distanceP,x);

> xcoord:=solve(distanceP=0);

> subs(x=xcoord,distanceDP);

=> The distance is a minimum (second derivative test) when x = 1. Therefore, the closest points are (1, ) and (1, ).

Note: To do this by hand , you would minimize the square of the distance function. Why??

e.g. Minimize cost

> restart:

Suppose a box with a square base that is to have a volume of 3 is to be made from three materials. The bottom is to be made from material that costs $10/ , the top from material that costs $1/ and the sides from material that costs $5/ . Determine the dimensions that will minimize the costs.

Let x = the height of the box. (The width will then be .)

> cost:=11*(3/x)+5*x*sqrt(3/x);

> plot(cost,x=0..10,y=0..100,colour=black);

> cost_prime:=diff(cost,x);cost_doubleprime:=diff(cost_prime,x);

> crit_number:=solve(cost_prime=0);

> evalf(crit_number,2);

> subs(x=crit_number,cost_doubleprime);

> evalf(%);

=> Cost is a minimum when x is about 3.8 meters (by the second derivative test).

> width:=sqrt(3/crit_number);evalf(%);

The dimensions should be 0.85 m by 0.88 m. to minimize the cost.

e.g. Maximize an area

Find the dimensions of the largest rectangle that can be inscribed inside a circle of radius r.

> with(plots):area:=4*x*sqrt(r^2-x^2);areaP:=diff(area,x);areaDP:=diff(areaP,x);

> plot3d(area,x=0..10,r=0..10);

> cnumbers:=solve(areaP=0,x);

> subs(x=cnumbers[1],areaDP);simplify(%);

It should be clear from at least one of the above that yields a maximum value for the area function.

Let's try the same problem again except now with a circle of radius 1 unit:

> area1:=subs(r=1,area);

> plot(area1,x=-1..1,y=-2..2,thickness=2);

> area1_prime:=diff(area1,x);

> cnumbs:=solve(area1_prime=0);

> subs(x=cnumbs[1],area1);

> subs(x=cnumbs[1],diff(area1_prime,x));