The question is: use second order Taylor polynomials for the given function about the point specified to approx the indicated value. Estimate the error and write the smallest interval you can be sure contains the value.

$f(x)=x^{1/3}$ about 8; approx $9^{1/3} $.

So I get that the approx about 8 is: $ 2 + \frac 1 {12}(x-8) (- 1 / (9 \times 32))\cdot(x-8)^2$

And when plugging in 9 i get that it's approximately equal to $ 2.07986 $.

When counting the error I use: $\left(\frac{f'''(c)}{3!}\right) \cdot (9-8)^3$ Since we want the error to be the maximum it can be I use $ c = 8 $. This gives that the error is maximum $0.000241$.

However, now comes the problem. I now want the interval that contains the value. I get that $9^{1/3} \text{ is } \leq 2.07986 + 0.000241 $ since it's obviously going to be equal or smaller than the approximation plus the maximum error.

But, I don't know how to get what value it's going to be bigger than. I would guess that's it going to be bigger than or equal to $2.07986 - 0.000241 $ since it can't be smaller than the approximation minus the maximum error to that. However, that's wrong when I check the solutions manual which states that $ 2.07986 < 9^{1/3} < 2.07986 + 0.000241 $.

So, two main questions:

  1. Shouldn't it be equal OR less/more than in the interval? Not just bigger or smaller than.

  2. Why don't you take that $ 9^{1/3} $ is bigger than the approximation MINUS the maxmimum error?

  • $\begingroup$ Well, $ 2.07986 < 9^{1/3} $ is not necessarily the same as $ 2.07986−0.000241< 9^{1/3} $. Why don't you subtract the maximum error from it's "bigger-than-value"? $\endgroup$ – gbgult Feb 11 '18 at 13:16
  • $\begingroup$ Yeah, but I don't get WHY the first one holds. We know that there's a maximum error of $0.000241$, why don't I have to take that into consideration with that inequality? $\endgroup$ – gbgult Feb 11 '18 at 13:21
  • $\begingroup$ Now I understand. The estimation of the error is the worst possible scenario, the actual error can be much smaller. $\endgroup$ – Peter Feb 11 '18 at 13:22
  • $\begingroup$ Yeah, I get that. But how do we know that $9^{1/3}$ is bigger than $2.07986$? Couldn't it be $9^{1/3} = 2.07985$ for example since we know that there's a potential error of maximum $0.000241$? $\endgroup$ – gbgult Feb 11 '18 at 13:29
  • $\begingroup$ Which remainder do you use ? The Lagrange-remainder ? $\endgroup$ – Peter Feb 11 '18 at 13:36

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