Asked by Lucy
Approximate the greatest real zero of the function g(x)= x^3-3x+1 to the nearest tenth.
I know that there is a zero between -2 and -1, and another between 0 and 1 but do not know how to find it to the nearest tenth. The only example shown in my book uses a calculator and mine does not have instructions for the same functions as the one in the book. I have checked for online calculators and cannot find one to do the calculations either.
Any help would be great as I have to get finished with math this week.
Thanks
I know that there is a zero between -2 and -1, and another between 0 and 1 but do not know how to find it to the nearest tenth. The only example shown in my book uses a calculator and mine does not have instructions for the same functions as the one in the book. I have checked for online calculators and cannot find one to do the calculations either.
Any help would be great as I have to get finished with math this week.
Thanks
Answers
Answered by
David Q
Try solving it iteratively: if g(X)=0, then X^3-3X+1=0, so rearrange the equation to read
X = (3X-1)^(1/3)
Not put X = 2, and evaluate the function. You'll get about 1.71. Feed that into the equation again, and you'll get about 1.60. Keep going for a few more iterations until it settles down. Then try feeding it into the original equation and see if it works.
X = (3X-1)^(1/3)
Not put X = 2, and evaluate the function. You'll get about 1.71. Feed that into the equation again, and you'll get about 1.60. Keep going for a few more iterations until it settles down. Then try feeding it into the original equation and see if it works.
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