Intersect the intervals after using partial fractions

Both copies print this first stage. Set

$\frac{3x+2}{(x-1)(x+4)}=\frac A{x-1}+\frac B{x+4},\qquad 3x+2=A(x+4)+B(x-1).$

At x=1, $5=5A$, so A=1. At x=−4, $-10=-5B$, so B=2. Hence $f(x)=1/(x-1)+2/(x+4)$.

The Student leaves the subsequent expansion blank; the Teacher completes

$\begin{aligned}&f(x)\\ & =(x-1)^{-1}+2(x+4)^{-1}\\ & =(-1)^{-1}(1-x)^{-1}\\ &{}+2(4)^{-1}(1+x/4)^{-1}.\end{aligned}$

Its full binomial substitution is

$\begin{aligned}&f(x)\\ & =-(1+x+x^2+x^3+\cdots)\\ &{}+\frac12\left[1+(-1)\frac x4+\frac{(-1)(-2)}{2!}\left(\frac x4\right)^2+\frac{(-1)(-2)(-3)}{3!}\left(\frac x4\right)^3+\cdots\right].\end{aligned}$

The Teacher writes

$\begin{aligned}&f(x)\\ & =-1-x-x^2-x^3+\frac12-\frac x8\\ &{}+\frac{x^2}{32}-\frac{x^3}{128}+\cdots\end{aligned}$

$=-\frac12-\frac98x-\frac{31}{32}x^2-\frac{129}{128}x^3+\cdots.$

The constant is the first term, so four terms run through $x^3$, not $x^4$.

For $(1-x)^{-1}$, the Teacher gives $|-x|<1\Rightarrow |x|<1\Rightarrow -1<x<1$. For $(1+x/4)^{-1}$, $|x/4|<1\Rightarrow |x|<4\Rightarrow -4<x<4$. The required intersection is $-1<x<1$. Both copies explicitly stress AND and INTERSECTION: satisfying either interval alone is insufficient. Independently, the rational function also exists outside this interval except at its poles; that does not extend this Maclaurin representation.