# The coordinates of the inflection point of the curve, y = e cos x , 0 ≤ x ≤ π .

### 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 29E
To determine

## To calculate: The coordinates of the inflection point of the curve,  y=ecosx,0≤x≤π .

Expert Solution

The infection point with six decimal place is x=0.904541 .

### Explanation of Solution

Given information:

The curve is given as:

y=ecosx,0xπ .

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.

Inflection point: Points can be find by f''(x)=0

Calculation:

Consider the curve ,

y=ecosx,0xπ

Now,

f(x)=ecosxf'(x)=ecosx(sinx)

=(sinx)ecosx

Differentiate f'(x)

f''(x)=sinx[ecosx(sinx)]+ecosx[(cosx)] (by product rule)

=(sinx)2ecosxcosxecosx

Hence, we get the inflection point by f''(x)=0 . In other words f''(x) is g(x) used to find out roots using Newton’s Method.

Therefore,

g(x)=(sinx)2ecosxcosxecosx and

g'(x)={(sinx)2[ecosx(sinx)]+ecosx[2sinxcosx]}{cosx[ecosx(sinx)]+ecosx[sinx]}

=(sinx)3ecosx+sin2xecosx+sinxcosxecosx+sinxecosx

Now, let initial approximation be x1=0.9

For n=1

x2=x1g(x1)g'(x1)

x2=x1(sinx)2ecosxcosxecosx(sinx)3ecosx+sin2xecosx+sinxcosxecosx+sinxecosx

x2=(0.9)(sin0.9)2ecos(0.9)cos(0.9)ecos(0.9)(sin0.9)3ecos(0.9)+sin2(0.9)ecos(0.9)+sin(0.9)cos(0.9)ecos(0.9)+sin(0.9)ecos(0.9)x2=(0.9)1.1424780671.1573900630.894933813+1.813229563+0.906614781+1.458494599

x2=0.9(0.014911996)3.28340513x2=0.9(0.004541625358)x2=0.904541625

The second approximation is x2=0.904541625 .

We establish the condition f''(x)0,f''(x)<0,f''(x+)>0 where . x=0.904541625

For =0.1

f''(0.9045416250.1)=f''(0.804541625)f''(0.804541625)=[sin(0.804541625)]2ecos(0.804541625)cos(0.804541625)ecos(0.804541625)

=1.0385832761.387291494=0.348708217<0

f''(0.904541625+0.1)=f''(1.004541625)f''(1.004541625)=[sin(1.004541625)]2ecos(1.004541625)cos(1.004541625)ecos(1.004541625)

=1.2178302870.917355663=0.300474624>0

Therefore, the inflection point of curve y=ecosx,0xπ is x=0.904541625 .

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