# The roots of the equation, e x = 3 − 2 x . ### 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 4.7, Problem 14E
To determine

## To calculate: The roots of the equation,  ex=3−2x .

Expert Solution

The roots are

x20.950212

x30.626051

x40.594447

### Explanation of Solution

Given information:

The equation is given as:

ex=32x .

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

xn+1=xnf(xn)f'(xn) and so on.

Calculation:

Consider the equation,

ex=32x

Now,

ex+2x3=0f(x)=ex+2x3f'(x)=ex+2

Let the initial approximation x1=1

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1ex1+2x13ex1+2

x2=(1)(e(1)+2(1)3e(1)+2)x2=(1)(e+23e+2)x2=(1)(e1e+2)

Value of e=2.063814508

x2=1(0.3678794417.389056099)x2=10.049787068x2=0.950212932

The second approximation is x2=0.950212 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2ex2+2x23ex2+2

x3=(0.950212)(e(0.950212)+2(0.950212)3e(0.950212)+2)x3=(0.950212)(2.586257888+1.90042432.586257888+2)x3=0.950212(1.4866818884.586257888)

x3=0.9502120.324160115x3=0.626051884

The third approximation with six decimal places is x3=0.626051 .

Let n=3 ,

x4=x3f(x3)f'(x3)

x4=x3ex3+2x33ex3+2

x4=(0.626051)(e(0.626051)+2(0.626051)3e(0.626051)+2)x4=(0.626051)(1.870210516+1.25210231.870210516+2)x4=0.626051(0.1223125163.870210516)

x4=0.6260510.031603582x4=0.594447418

The fourth approximation with six decimal places is x4=0.594447 .

Therefore, the roots with six decimal places are :-

x20.950212

x30.626051

x40.594447

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