1

.1x

xh(x) 2

3

Example 3 Sketch the graph of the function

Solution Observe that h is an odd function, and its graph is symmetric with respect to the origin.I. InterceptsThe x-intercepts occur when 0 = x3 , i.e. when x=0.

The y-intercept occurs at h(0)=0.

II. Asymptotes Vertical asymptotes occur where the denominator of h(x) is zero: 0 = x2-1 = (x+1)(x-1), i.e. when x=-1 and x=1.Since the degree 3 of the numerator of h is one larger than the degree 2 of the denominator, the graph of h has an oblique asymptote which we can find by long division.

Hence and the graph of h has the line y=x

as an oblique asymptote on both the left and the right.

xxx

xx1x

3

32

,1x

xx1x

xh(x) 22

3

2

III. First Derivative

By the quotient rule, the derivative of is:

Since the denominator of h /(x) is always positive, h /(x) has the same sign as its numerator. Since x2 0, h /(x) has the same sign as Hence h

/(x) is positive for while h /(x) is negative for Thus h is increasing for while h is decreasing for We depict this information on a number line.

h has three critical points: at x=0, where the numerator of h /(x) vanishes. By the First Derivative Test, is a local maximum, x=0 is not a local extremum and is a local minimum. Note that x=-1 and x=1 are not critical points of h because h has vertical asymptotes at these numbers and they are not in the domain of h.

22

2

22

22

22

24

22

424

22

322

1x1x3x3xx

1x3xx

1xx3x

1xx2x3x3

1xx2x1xx3xh )()(

))(()(

)()()(

)()()(

i n cr -1 0 1 i n cr

si gn h’(x)x

+ + + 0 - - - - - - - - - - 0 - - - - - - - - 0 + + +

loca lm ax

loca lm i n

decr easi n g decr easi n g

1xxh(x) 2

3

).)(( 3x3x 3x3x or .3x3

3x3x or .3x3

3x3x and 3x

3x

3 3

3

IV. Vertical Tangents and Cusps

h has neither vertical tangents nor cusps.

V. Concavity and Inflection Points

By the quotient rule, the derivative of is:

Observe that x2+3 in the numerator of h //(x) is always positive. Hence h //(x) is positive for –1<x<0 or x>1 and negative for x<-1 or 0<x<1. Therefore the graph of h is concave up for –1<x<0 or x>1 and concave down for x<-1 or 0<x<1. Note that h has vertical asymptotes x=-1 and x=1, and these numbers are not in the domain of h. Hence h has only one inflection point x=0 where the concavity changes from up to down.

con c dow n -1 con c u p 0 con c dow n 1 con c u px

si gn h“(x )- - - - - - + + + 0 - - - - - - + + + +

i n fl ecti onpoi n t

22

24

1xx3xxh )()(

33

2

33

3

33

3535

32

2423

42

224223

1x1x3xx2

1x1xx6x2

1x1xx12x4x6x10x4

1xx4x3x1xx6x4

1xx21x2x3x1xx6x4xh

)()()(

)()()()()(

)())(())((

)())(()())(()(

4

VI. Sketch of the graph

We summarize our conclusions and sketch the graph of h.

x-intercepts: 0 y-intercept: 0 vertical asymptotes: x=-1 and x=+1

oblique asymptote: y=x on the left and right increasing:

decreasing: local max: local min:

concave up: -1<x<0 or x>1 concave down: x<-1 or 0<x<1

inflection point: x=0 h is an odd function

3x3x or 3x3

3

233,

1xxh(x) 2

3

3

233,

3

3