Asked by TayB
Use Newton's method with initial approximation
x1 = −2
to find x2, the second approximation to the root of the equation
x^3 + x + 1 = 0.
*I got -1.307692 for my answer but it said that was wrong, so then I tried rounding it to two decimal places and got -1.31 and that still didn't work and then i tried -1.30 and it still didn't work
x1 = −2
to find x2, the second approximation to the root of the equation
x^3 + x + 1 = 0.
*I got -1.307692 for my answer but it said that was wrong, so then I tried rounding it to two decimal places and got -1.31 and that still didn't work and then i tried -1.30 and it still didn't work
Answers
Answered by
Steve
Visit
http://keisan.casio.com/exec/system/1244946907
and type in your functions. You will see that the root is -0.68233
You can also see this at wolframalpha.com
When you try to solve a problem and don't get a satisfactory answer, it would help if you showed your work. Then we could see where you went astray.
http://keisan.casio.com/exec/system/1244946907
and type in your functions. You will see that the root is -0.68233
You can also see this at wolframalpha.com
When you try to solve a problem and don't get a satisfactory answer, it would help if you showed your work. Then we could see where you went astray.
Answered by
TayB
-0.68233 is still wrong. I had tried that one earlier too
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