Use the second order Taylor polynomials $P_2(x)$ for the given functions about the point specified to approximate the indicated value. Estimate the error and write the smallest interval you can be sure contains the value. Calculator not allowed.
Well I know the formula
$P_2(x)=f(a)+f'(a)(x-a)+\frac{f''(x)}{2}(x-a)^2$
But as soon as I try to solve any equations everything just become wrong and impossible to solve without calculator and even with calculator I get the wrong answer.
$f(x)=x^{1/3}$ about 8; approximate $9^{1/3}$
$f(x)=\frac{1}{x} $ about 1; approximate $\frac{1}{1.02}$
If I understood it correctly:
$1: a=8\quad x=9^{1/3}$
$2:a=1\quad x=\frac{1}{1.02}$
$\endgroup$ 11 Answer
$\begingroup$If $f(x)=\sqrt[3]x$ and $a=8$, then$$P_2(x)=2+\frac{x-8}{12}-\frac{(x-8)^2}{288}.$$ Therefore $P(9)=\frac{599}{288}$. The error has the form $\frac{f^{(3)}(\zeta)}{3!}(9-8)^2=\frac{f^{(3)}(\zeta)}6$, for some $\zeta\in[8,9]$. Since $f^{(3)}(x)=\frac{10x^{-8/3}}{27}$, which is a decreasing function, $\frac{f^{(3)}(\zeta)}6\leqslant\frac{10\times8^{-8/3}}{6\times27}=\frac 5{20\,736}\simeq0.00024$.
Can you deal with the other exercise now?
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