Friday

October 31, 2014

October 31, 2014

Posted by **Kelly** on Thursday, February 3, 2011 at 12:40am.

2x^3-2x^2-2x+1=0

i am completely lost & have no idea where to start.

- calculus -
**drwls**, Thursday, February 3, 2011 at 6:30amFind two values of x that result a positive and negative value for the cubic polynomial on the left. If you can do that, there must be an intermediate value of x that results in zero.

If x=0, 2x^3-2x^2-2x+1 = 1

If x=1, 2x^3-2x^2-2x+1 = -1

So there must be a solution between x = 0 and x = 1

I do not have a graphing calculator to help you with the second part, but you can always plot the graph by hand and interpolate.

For x = 0.5, f(x) = -0.25

For x = 0.4, f(x) = 0.008

For x = 0.404, f(x) = -0.0026

So there is a solution around x = 0.403

For a more exact answer, see http://www.1728.com/cubic.htm

There are actually three roots to the equation. In just solved for one of them

0.40303171676268..

-0.85463767971846.., and

1.4516059629557..

The other two could be found by taking initial x assumed values above and below the roots shown (e.g. 0 and -1, or 1 and 2), and interpolating, as I did for 0 and 1.

This is not calculus, by the way.

**Answer this Question**

**Related Questions**

Math Calculus - The Image Theorem: The image theorem, a corollary of the ...

calculus - Use the Intermediate Value Theorem to show that there is a root in ...

calculus - use the fundamental theorem to find the area under f(x)=x^2 between x...

calculus - "use the intermediate value theorem to prove that the curves y=x^2 ...

Calculus (Please Check) - Show that the equation x^5+x+1 = 0 has exactly one ...

precalculus - Use the Intermediate Value Theorem and a graphing utility to find...

algebra2 - Plz Explain how to use a graphing calculator to solve the equation: x...

calculus - Verify that the Intermediate Value theorem applies to the indicated ...

Algebra 2 - I need help with this problem because I do not have a graphing ...

Math - Use the Intermediate Value Theorem to show that there is a root of the ...