# The function f ( x ) = x 2 − 2 x 1000 at the value of x = 1 , 0.8 , 0.6 , 0.4 , 0.2 , 0.1 and 0 .05

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 2.2, Problem 29E

(a)

To determine

## To find:The function f(x)=x2−2x1000 at the value of x=1,0.8,0.6,0.4,0.2,0.1 and 0.05

Expert Solution

f(x)=0.001

### Explanation of Solution

Given:

The function f(x)=limx0(x22x1000)

Concept used:

The left derivative and right derivative of a function f at a point x=c are , equal

f'(c)=limh0f(ch)f(c)hf+'(c)=limh0+f(c+h)f(c)h

Calculation:

The function

y=f(x)=x22x1000.............(1)x=1f(1)=12211000=100021000=9981000=0.998x=0.8f(0.8)=(0.8)22(0.8)1000=0.6382x=0.6f(0.6)=(0.6)22(0.6)1000=0.3584x=0.4f(0.4)=(0.4)22(0.4)1000=0.1586x=0.2f(0.2)=(0.2)22(0.2)1000=0.0588x=0.1f(0.1)=(0.1)22(0.1)1000=0.0089x=0.05f(0.05)=(0.05)22(0.05)1000=0.0014

f(x)=limx0(x22x1000)f(x)=011000f(x)=0.001

(b)

To determine

Expert Solution

f(x)=0.001

### Explanation of Solution

Given:

The function f(x)=limx0(x22x1000)

Concept used:

The left derivative and right derivative of a function f at a point x=c are , equal

f'(c)=limh0f(ch)f(c)hf+'(c)=limh0+f(c+h)f(c)h

Calculation:

The function

y=f(x)=x22x1000.............(1)x=0.04f(0.04)=(0.04)22(0.04)1000=0.00057x=0.02f(0.02)=(0.02)22(0.02)1000=0.00061x=0.01f(0.01)=(0.01)22(0.01)1000=0.00091x=0.005f(0.005)=(0.005)22(0.005)1000=0.00097x=0.003f(0.003)=(0.003)22(0.003)1000=0.00091x=0.001f(0.001)=(0.001)22(0.001)1000=0.000990

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