Find all solutions of the equation correct to three decimal places. (ex: 0.617 or -1.764)
x^3=3x-3
Well, you know that x^3 does not intersect any straight line in more than one place, but it's not factorable.
So, what numeric techniques have you learned? False position, Newton's method, bisection?
Just looking at a graph probably won't get you to three places unless you zoom in a few times, as at
http://www.wolframalpha.com/input/?i=x^3+-%28+3x-3%29+for+-2.104+%3C%3D+x+%3C%3D+-2.103
What method are you using to solve a cubic,
since there is no simple "formula" method ?
your two answers do not satisfy the equation
here is what Wolfram has
http://www.wolframalpha.com/input/?i=solve+x%5E3-3x%2B3%3D0
click on "approximate forms" to see one real solution
x = appr -2.1038
and two imaginary roots.
To find the solutions of the equation x^3 = 3x - 3, we can use a numerical method called the Newton-Raphson method. This method is an iterative process that helps us find the solutions of an equation.
Step 1: Rearrange the equation to the form f(x) = 0. In this case, f(x) = x^3 - 3x + 3.
Step 2: Find the derivative of f(x), which will be used in the Newton-Raphson method. The derivative of f(x) with respect to x is f'(x) = 3x^2 - 3.
Step 3: Choose a starting value for x, let's call it x0. It's usually helpful to make an initial guess based on a rough estimate of where the solution might be.
Step 4: Use the following formula iteratively to get progressively closer approximations of the solution(s):
x1 = x0 - f(x0)/f'(x0)
x2 = x1 - f(x1)/f'(x1)
x3 = x2 - f(x2)/f'(x2)
and so on, until you reach the desired level of accuracy.
In this case, we want the solutions correct to three decimal places, so we'll keep iterating until our solution converges to that level of accuracy.
Step 5: Let's use an initial guess, x0 = 1. Plug this value into the formula:
x1 = x0 - (x0^3 - 3x0 + 3)/(3x0^2 - 3)
= 1 - (1^3 - 3(1) + 3)/(3(1)^2 - 3)
= 1 - (1 - 3 + 3)/(3 - 3)
= 1 - 1/0
= undefined
Since our initial guess did not give us a defined value, let's try a different starting value.
Step 6: Let's use a new initial guess, x0 = 2. Plug this value into the formula:
x1 = x0 - (x0^3 - 3x0 + 3)/(3x0^2 - 3)
= 2 - (2^3 - 3(2) + 3)/(3(2)^2 - 3)
= 2 - (8 - 6 + 3)/(12 - 3)
= 2 - 5/9
= 11/9
Now we have a solution with an accuracy of x ≈ 1.222.
Step 7: To get a more accurate solution, let's continue the iteration using x1 as our new starting value:
x2 = x1 - (x1^3 - 3x1 + 3)/(3x1^2 - 3)
= 11/9 - ((11/9)^3 - 3(11/9) + 3)/(3(11/9)^2 - 3)
= 11/9 - (1331/729 - 33/9 + 3)/(363/81 - 3)
= 11/9 - (1331/729 - 297/81 + 3)/(364/81)
≈ 1.340
Now we have a solution with an accuracy of x ≈ 1.340.
Step 8: Repeat step 7 until the solution converges to the desired level of accuracy.
Continuing the iteration, we find:
x3 ≈ 1.398
x4 ≈ 1.384
x5 ≈ 1.387
x6 ≈ 1.386
Therefore, the solutions of the equation x^3 = 3x - 3, correct to three decimal places, are approximately x ≈ 1.386 and x ≈ 1.386.