Mr. Pursley has answered both questions.
If you had read the links, you would be able to come up with preliminary, if not final answers.
If you encounter difficulties, post in detail what they are.
Can someone help me out here with these two problem? I have know idea where to start.
1)The equation: 10(x-1)(x-2)(x-3)=1,
has three real solutions a<b<c
where
a=____, b=___ , and c=___.
Enter your answers with at least six correct digits beyond the decimal point.
Hint: Ask what the solutions are if the right hand side is 0 instead of 1, and use Newton's Method.
2) Find the smallest positive value of which satisfies -- x=4.100cos(2.800x)
Give the answer to four places of accuracy._________
Remember to calculate the trig functions in radian mode.
2 answers
I'll give you a headstart.
Rewrite the equation in the form:
f(x)=0
For example
f(x)=x²-2=0
Calculate f'(x)=2x
Start with an approximation, say x0=1.
Calculate
x1=x0-f(x0)/f'(x0)
=1-(1-2)/2
=1.5
Repeat until the desired accuracy is obtained:
x2=x1-f(x1)/f'(x1)
=1.5-(2.25-2)/(2*2.25)
=1.444
x3=x2-f(x2)/f'(x2)
=1.444-(0.085136)/2.888
=1.41452
...
Notice that the number of accurate figures of the solution doubles with every iteration.
Try the method with the given problems and post if you have difficulties.
Rewrite the equation in the form:
f(x)=0
For example
f(x)=x²-2=0
Calculate f'(x)=2x
Start with an approximation, say x0=1.
Calculate
x1=x0-f(x0)/f'(x0)
=1-(1-2)/2
=1.5
Repeat until the desired accuracy is obtained:
x2=x1-f(x1)/f'(x1)
=1.5-(2.25-2)/(2*2.25)
=1.444
x3=x2-f(x2)/f'(x2)
=1.444-(0.085136)/2.888
=1.41452
...
Notice that the number of accurate figures of the solution doubles with every iteration.
Try the method with the given problems and post if you have difficulties.