Sketching (x - 1)/(x^2 - 4)


As preliminary work we name our function g(x) and compute g'(x), g''(x), and a few values for g:

x
g(x) =
(x2 - 4)


1*(x2 - 4) - x*(2x) - (x2 + 4)
g'(x) =
=
(x2 - 4)2 (x2 - 4)2


(-2x)*(x2 - 4)2 - ( -(x2 + 4) )*( 2(x2 - 4)(2x) )
g''(x) =
=
(x2 - 4)4
(-2x)(x2 - 4) +(x2 + 4)(4x) (2x)(x2 + 12)

=
(x2 - 4)3 (x2 - 4)3

We now go down the list:

Vertical Asymptotes. There are vertical asymptotes at x = ± 2 because the denominator is zero there and the numerator is not.

To sketch the curve we will need to know the nature of these asymptotes. Specifically, does the function go to or - as x approaches -2 and 2 from the left and from the right.

Horizontal Asymptotes. Since the degree of the denominator is greater than that of the numerator, g(x) approaches 0 as |x| gets large. Thus y = 0 is a horizontal asymptote.

To sketch the curve we will need to know if it approaches the line x = 0 from below or from above.

Symmetry. We note that g(-x) = -g(x). Thus g is an odd function and the graph is symmetric about the line y = -x.

Continuity. The function and its derivatives are continuous everywhere except at x =± 2.

Critical Points. f(x) is differentiable for x ± 2. This domain contains no endpoints. Hence critical points other than x = ± 2 can only occur at places where the derivative is zero. Since the derivative is never zero, there are no other critical points (and hence no extreme points).

The only obvious value of the function is g(0) = 0. So at this point we have:

The concavity of the graph is governed by the sign of g'(x). The numerator of g'(x) is positive if x is positive and negative if x is negative. It is zero at x = 0 and changes sign there. The denominator is positive for |x| > 2, negative for |x| < 2. We can display this as follows:

Range of x-values: < - 2 -2 -2 to 000 to 2 2 > 2
Sign of Numerator: - - - 0+ ++
Sign of Denominator: + 0 - - - 0 +
Sign of g''(x): - DNE + 0 - DNE +

Thus the graph of g(x) is concave down for x < -2, concave up for -2 < x < 0, and so on. Since the concavity changes at x = 0, this is an inflection point.

Finally, we note that g(x) itself is negative for large negative numbers and positive for large positive numbers, which tells us on which side the graph approaches x = 0.

Putting this all together, the graph starts below x =0 and is concave down so must go to - to the left of x = -2. To the right of x = -2 the graph is concave up, so it must go to to the right of x = -2. By symmetry, or continuing with the information given by the second derivative, the graph goes to - to the left of x = 2 and to - to the right. With no critical points other than the places where g is not defined, there are no bumps or other features. We get the following sketch which is displayed next to a computer-generated graph.

Sketch of x/(x2 - 4)
Graph of x/(x2 - 4)

© CalculusQuestTM
Version 1
All rights reserved---1996
William A. Bogley
Robby Robson