Question
Use the Intermediate Value Theorem to prove that the equation has a solution. Then use a graphing calculator or computer grapher to solve the equation.
2x^3-2x^2-2x+1=0
i am completely lost & have no idea where to start.
2x^3-2x^2-2x+1=0
i am completely lost & have no idea where to start.
Answers
drwls
Find 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.
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.