CurveSketching.mws

Applications of Differentiation: Summary of Curve Sketching

Vocabulary

domain, intercept, symmetry, asymptote, critical number, concavity, inflection point

Objective

Sketch accurate graphs of functions, identifying important properties which have been discussed in the course up to this point.

Lecture Outline

Thus far we have discussed many properties of functions. We list some of the most important of these properties below.

Properties of Functions

Properties one through four in the list above can often be determined without the use of calculus techniques while properties five through seven require consideration of the first and second derivatives of the function.

Given any function, we can methodically determine each of the properties and then sketch a graph of a function satisfying those properties. The sketch obtained in this way is often a reasonably accurate representation of the graph of the initial function.

Example: f(x) = x^2/(x^2-9)

We methodically determine each of the properties in the list.

> f:=x->x^2/(x^2-9);

f := proc (x) options operator, arrow; x^2/(x^2-9) ...

The domain of f consists of all real numbers x such that the denominator is not equal to zero.

> solve(x^2-9,x);

3, -3

Hence the domain is all real numbers excluding 3 and -3.

The y-intercept is the point (0,f(0)) provided that 0 is in the domain of f.

> f(0);

0

We see that (0,0) is the y-intercept.

The x-intercepts occur wherever the numerator of f is equal to 0.

> solve(x^2,x);

0, 0

Hence there is only one x-intercept and it has x-coordinate 0. The x-intercept itself is the point (0,0).

Now we consider the symmetry of f.

> f(-x);

x^2/(x^2-9)

Note that f(-x)=f(x). We conclude that f is an even function (i.e., the graph of f is symmetric with respect to the y-axis).

Since f is a rational function in reduced form (i.e., the numerator and denominator share no common factors), the vertical asymptotes of f are obtained by finding where the denominator is equal to 0. We have already determined that the denominator is 0 when x=3 or x=-3 when we considered the domain of f. Hence the graph of f has two vertical asymptotes, the lines x=3 and x=-3.

The horizontal asymptote is found by determining Limit(x^2/(x^2-9),x = infinity)

> limit(x^2/(x^2-9),x=infinity);

1

We thus have that the line y=1 is the horizontal asymptote of the graph of f. Recall that a rational function has at most one horizontal asymptote, so we do not need to consider Limit(x^2/(x^2-9),x = -infinity) .

The intervals of increase and decrease may be determined using the increasing/decreasing test. We compute the first derivative of f and determine where it is positive or negative.

> fprime:=x->D(f)(x);

fprime := proc (x) options operator, arrow; 2*x/(x^...

> solve(fprime(x)>0,x);

RealRange(-infinity,Open(-3)), RealRange(Open(-3),O...

> solve(fprime(x)<0,x);

RealRange(Open(0),Open(3)), RealRange(Open(3),infin...

We conclude that f is increasing on the intervals ( -infinity ,-3) and (-3,0) while f is decreasing on the intervals (0,3) and (3, infinity ).

The critical numbers of f occur where the first derivative is equal to 0 or is undefined. Since f is a rational function, the only critical numbers where f can have local extrema are those for which the first derivative is equal to 0 (recall that these are also called stationary numbers). We solve for those critical numbers.

> solve(fprime(x),x);

0

There is only one stationary number, x=0. We can classify this critical number as a local extremum using the first derivative test. The first derivative changes sign from positive to negative across 0, so f has a local maximum at x=0.

The concavity of f is determined by using the concavity test.

> fdoubleprime:=x->D(fprime)(x);

fdoubleprime := proc (x) options operator, arrow; 2...

> solve(fdoubleprime(x)>0,x);

RealRange(-infinity,Open(-3)), RealRange(Open(3),in...

> solve(fdoubleprime(x)<0,x);

RealRange(Open(-3),Open(3))

We conclude that the graph of f is concave upward on the intervals ( -infinity ,-3) and (3, infinity ) while the graph of f is concave downward on the interval (-3,3).

Inflection points occur at points on the graph where concavity changes direction. In this case, it is apparent that concavity changes across x=-3 and x=3, but these are not in the domain of f, so there are no inflection points.

Let us verify the information about the properties of f we have obtained above by comparison with a Maple plot of the graph of f.

> plot(f(x),x=-5..5,y=-20..20);

[Maple Plot]

Exercise: g(x) = x/(1+x^2)

Use the same process as above to determine the important properties of the graph of g, make a sketch of a function having those properties, and then compare this sketch with a Maple plot of the graph of g.

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