Chapter 4, Problem 2PS

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Relative Extrema(a) Graph the fourth-degree polynomial f ( x ) = a x 4 − 6 x 2 for a = − 3 , − 2 , − 1 , 0 , 1 , 2 , and 3. For what values of the constant a does p have a relative minimum or relative maximum?(b) Show that p has a relative maximum for all values of the constant a.(c) Determine analytically the values of a for which p has a relative minimum.(d) Let ( x , y ) = ( x , p ( x ) ) be a relative extremum of p. Show that (x, y) lies on the graph of y = − 3 x 2 . Verify this result graphically by graphing y = − 3 x 2 together with the seven curves from part (a).

(a)

To determine

To graph: The function p(x)=ax46x2 for different integer values of a from 3 to 3 inclusive and to determine the values of a for which the function p has relative maxima or minima.

Explanation

Graph:

Consider the function when a takes values 0, 1, 2 and 3.

p0(x)=âˆ’6x2p1(x)=x4âˆ’6x2p2(x)=2x4âˆ’6x2p3(x)=3x4âˆ’6x2

Now consider the function when a takes values âˆ’1,âˆ’2,âˆ’3.

pâˆ’1(x)=âˆ’x4âˆ’6x2pâˆ’2(

(b)

To determine

To prove: The function p(x)=ax46x2 has a relative maximum for all a.

(c)

To determine

To calculate: The values of a such that the provided function p(x)=ax46x2 has only 1 relative minimum.

(d)

To determine

To prove: The relative extrema for the function p(x)=ax46x2 lies on the curve y=3x2.

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