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Chapter 4: Q. 68 (page 242)

For what positive numbers will the cube of a number be less than the number?

Short Answer

Expert verified

The cube of a number will be less than the number in the set of{x|0<x<1}.

Step by step solution

01

Step 1. Given Information 

We have to find out for what positive numbers will the cube of a number be less than the number.

02

Step 2. Form an inequality expression 

To find the positive numbers, let the number bex.

According to the question,

x3<xx3-x<0

03

Step 3. Solve the inequality expression 

Let f(x)=x3-x

Now, let's find the real zeros

x3-x=0xx2-1=0x=0andx2-1=0x+1x-1=0x=1,-1

04

Step 4. Check the points 

From the real zeros, we get three intervals which are -鈭�,-1,-1,0,0,1,1,鈭�.

Let x=-3, substitute in fx=x3-x

f-3=-33--3f-3=-27+3f(-3)=-24

It is negative.

Let x=-12, substitute in fx=x3-x

f-12=-123--12f-12=38

It is positive.

Let x=12, substitute in fx=x3-x

f12=123-12f12=-38

It is negative.

Let x=3,substitute in fx=x3-x

f3=33-3f3=27-3f3=24

It is positive.

Thus, the set is -鈭�,-1and0,1becausefx<0but we need positive numbers the solution will be{x|0<x<1}.

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