Posted by **REALLY NEED HELP!!!!** on Thursday, November 4, 2010 at 5:12pm.

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.

- Math: Calculus -
**MathMate**, Thursday, November 4, 2010 at 5:21pm
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.

- Math: Calculus -
**MathMate**, Thursday, November 4, 2010 at 6:15pm
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.

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