Question
Asked Nov 29, 2019
36 views

Use newtons method to approximate a root of the equation e^(-x) = 5 + x correct to eight decimal places.

check_circle

Expert Answer

Step 1

Use newtons method to approximate a root of the equation e(-x) = 5 + x correct to eight decimal places.

Step 2

Consider the equation

e-x=5+x

Rewrite the aquation
-5-x-0
Differanticate the fiunction by xaz follows
f'(x)= --1
equate f'(x)=0
--1-0
xine-In(-1)
x-3.14
choose the irnitial value as follows
x15
By Newton's method
f(x.)
f (x,)
f'(%)
1.5--15)
f(-1.5)
f(-1.5)-0.981689070
f(-1.5)--5.418689070
Therefore
0.981689070
x--15-
(-5.481689070)
--1.320914857
help_outline

Image Transcriptionclose

Rewrite the aquation -5-x-0 Differanticate the fiunction by xaz follows f'(x)= --1 equate f'(x)=0 --1-0 xine-In(-1) x-3.14 choose the irnitial value as follows x15 By Newton's method f(x.) f (x,) f'(%) 1.5--15) f(-1.5) f(-1.5)-0.981689070 f(-1.5)--5.418689070 Therefore 0.981689070 x--15- (-5.481689070) --1.320914857

fullscreen
Step 3

At n=1

 ...

By Newton's method
f(x,)
2
f'(x)
f(-1.320914857)
f'(-1.320914857)
=-1.320914857
f(-1.320914857) = 0.067762497
f'(-1.320914857) = -4.746847640
Therefore,
0.067762497
x -1.320914857 -
(-4.746847640)
-1.306639594
help_outline

Image Transcriptionclose

By Newton's method f(x,) 2 f'(x) f(-1.320914857) f'(-1.320914857) =-1.320914857 f(-1.320914857) = 0.067762497 f'(-1.320914857) = -4.746847640 Therefore, 0.067762497 x -1.320914857 - (-4.746847640) -1.306639594

fullscreen

Want to see the full answer?

See Solution

Check out a sample Q&A here.

Want to see this answer and more?

Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*

See Solution
*Response times may vary by subject and question.
Tagged in

Math

Calculus