Choose a valid substitution and assess the resulting estimate

The Teacher supplies the normalization left blank in the Student copy. Let $F(x)=\sqrt[3]{(x+2)/(x-1)}$. Using real cube roots near zero,

$\begin{aligned}&F(x)\\ & =(x+2)^{1/3}(x-1)^{-1/3}\\ & =2^{1/3}(1+x/2)^{1/3}(-1)^{-1/3}(1-x)^{-1/3}.\end{aligned}$

Here $(-1)^{-1/3}=-1$, so the prefactor is $-\sqrt[3]2$. Both copies then print the two complete binomial brackets:

$\begin{aligned}&F(x)\\ & =-2^{1/3}\left[1+\frac13\frac x2+\frac{(1/3)(-2/3)}2\left(\frac x2\right)^2+\cdots\right]\end{aligned}$

$\hspace{1em}\times\left[1+\left(-\frac13\right)(-x)+\frac{(-1/3)(-4/3)}2(-x)^2+\cdots\right].$

Both copies simplify these brackets to

$-2^{1/3}\left(1+\frac x6-\frac{x^2}{36}+\cdots\right)\left(1+\frac x3+\frac{2x^2}{9}+\cdots\right).$

The quadratic coefficient includes the product of the two linear terms:

$F(x)\approx-2^{1/3}\left(1+\frac x6+\frac x3-\frac{x^2}{36}+\frac{2x^2}{9}+\frac{x^2}{18}\right)$

$=-2^{1/3}\left(1+\frac x2+\frac{x^2}{4}\right).$

This is the requested quadratic approximation; later terms begin at order $x^3$.

Both copies state $|x/2|<1\Rightarrow |x|<2\Rightarrow -2<x<2$ and $|-x|<1\Rightarrow |x|<1\Rightarrow -1<x<1$. Their intersection is $-1<x<1$.

The second interval line labels its factor $(1-x)^{1/3}$, but the preceding derivation uses $(1-x)^{-1/3}$. This is a source exponent-sign typo, preserved in the native text and corrected explicitly here. The stated sufficient interval remains $|x|<1$ for the factor actually being expanded.

Both copies print the substitution setup:

$F(8/11)=\sqrt[3]{-10}=-\sqrt[3]5\sqrt[3]2,$

$\begin{aligned}&-2^{1/3}\left[1+\frac12\left(\frac8{11}\right)+\frac14\left(\frac8{11}\right)^2\right]\\ & =-\sqrt[3]2\left(\frac{181}{121}\right).\end{aligned}$

The Teacher completes $-\sqrt[3]5\sqrt[3]2\approx-\sqrt[3]2(181/121)$, hence $\sqrt[3]5\approx181/121$. The printed decimal is 1.4958. Independently, $181/121=1.495867768\ldots$, so the correctly rounded estimate to four decimal places is 1.4959. This rounds the estimate; it does not certify accuracy to four decimal places relative to the true cube root.

The Student leaves this answer blank. The Teacher identifies Brian, because x=3 does not satisfy $-1<x<1$. Clare’s $x=-1/3$ lies inside that interval. No numerical evaluation is needed for this rejection.

Both copies print the second substitution setup:

$F(-1/3)=\sqrt[3]{-5/4}=-\frac{\sqrt[3]5}{\sqrt[3]4},$

$\begin{aligned}&-2^{1/3}\left[1+\frac12\left(-\frac13\right)+\frac14\left(-\frac13\right)^2\right]\\ & =-2^{1/3}\left(\frac{31}{36}\right).\end{aligned}$

The Teacher completes

$\begin{aligned}&\sqrt[3]5\approx\sqrt[3]4\sqrt[3]2\left(\frac{31}{36}\right)\\ & =\frac{31}{18}\\ & =1.722222\ldots\approx\boxed{1.7222}.\end{aligned}$

The Teacher says Clare’s estimate is better because $-1/3$ is closer to zero than $8/11$. Both copies remark that using the first few Maclaurin terms gives a better approximation when x is closer to zero. For a fixed expansion and truncation near zero this is a useful guide; it is not a universal monotonic-error theorem across all functions and signed inputs.

The source gives the true value approximately as 1.71. Independently, $\sqrt[3]5=1.709975946\ldots$; the absolute errors of $181/121$ and $31/18$ are about 0.214108 and 0.012246 respectively, confirming the improvement in this example. Both the first and second substitutions are inside the convergence interval, so convergence alone does not guarantee a good quadratic estimate.