   Chapter 4.3, Problem 86E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# For what values of c does the polynomial P(x) = x4 + cx3 + x2 have two inflection points? One inflection point? None? Illustrate by graphing P for several values of c. How does the graph change as c decreases?

To determine

To find: The value of c does the polynomial P(x)=x4+cx3+x2 with two inflection points, one inflection point and none of the inflection point.

Explanation

Calculation:

The given function is, P(x)=x4+cx3+x2 .

Obtain the derivative of P(x) .

P(x)=ddx(x4+cx3+x2)=4x3+3cx2+2x

Obtain the derivative of P(x) .

P(x)=ddx(4x3+3cx2+2x)=4(3x2)+3c(2x)+2(1)=12x2+6cx+2

Set P(x)=0 to check the nature of c.

12x2+6cx+2=0

The quadratic expression is, x=b±b24ac2a where a=12 , b=6c and c=2 .

x=(6c)±(6c)24(12)(2)2(12)x=6c±36c29624

Case (1):

If the discriminant is positive, then,

36c296>036c2>96c2>9636|c|>263

In this case if P(x) changes its sign twice

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