   Chapter 3.6, Problem 21E

Chapter
Section
Textbook Problem

# 20–25 Describe how the graph of f varies as c varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c . changes. You should also identify any transitional values of c at which the basic shape of the curve changes. f ( x ) = x 2 + 6 x + c / x (Trident of Newton)

To determine

To describe:

How the graph of fx  varies as c varies.

Explanation

1) Concept:

From the graph, identify the maximum and minimum points and points of inflection

2) Given:

fx=x2+6x+cx

3) Calculation:

We have

fx=x2+6x+cx

Draw the graph of fx,

From the graph,

Here, when c=0 then f(x) intersects x-axis in two points and

When c<0 then f(x) intersects x-axis in three points

When c>0 then f(x) intersects x-axis in only one point

Here three cases as; c=0, c<0  and c>0

Now find out maximum and minimum points,

Differentiate f(x) on both sides with respect to x, we get

f'x=2x+6-cx2

Again Differentiate on both sides with respect to x,

f''(x)=2+2cx3

If c=0 then graph of given function is of the form of parabola

That means there is only minimum value at -3, -9 (from the graph)

If c<0,

Draw the graph of f(x),

First solve fx for finding the x- intercepts

So, solve fx=0

x2+6x+cx=0

Consider, gx=x3+6x2-c

Assume that c=-32

x2+6x-32x=0

x3+6x2-32=0

By solving the equation by synthetic division method gives one factor,

x=-4

So consider gx has maximum at x=-4

g-4=-43+6-42-c

0=32+c

c<-32

Here if c<-32  the maximum is negative but there are no negative x- intercepts and one positive x- intercepts

So if the maximum is 0 then there are one negative x- interceptsand one positive x- intercepts

If -32<c<0, maximum is positive but there are two negative x- interceptsand one positive x- intercepts

As c0-, the minimum point approach to -3,-9

And fx has no maximum value for c<0

Now find out inflection points,

So, solve f''x=0

2+2cx3=0

-2x3=2c

x=-c3

So f-c3=-c32+6-c3+c-c3

=-c32+6-c3--c3-2

=6-c3

So the inflection point is, -c3, 6-c3

For c>0,

Draw the graph of fx,

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