which table shows a function that is increasing only over the interval (-2, 1), and nowhere else?

Answer:

The first - table (left - most table).

Explanation:

Step1: Define increasing function

A function is increasing when $x_1<x_2$ implies $f(x_1)<f(x_2)$.

Step2: Check first table

For $x$ in $(-2,1)$: when $x=-2,f(x)= - 3$; when $x=-1,f(x)=-1$; when $x = 0,f(x)=1$; when $x = 1,f(x)=3$. It is increasing in $(-2,1)$. For $x=-3,f(x)=-6$ and $-6<-3$ (when $x=-2$), and for $x = 2,f(x)=6$ and $3<6$, but it is only increasing in $(-2,1)$.

Step3: Check second table

In the interval $(-2,1)$, it is increasing. But for $x=-3,f(x)=-2$ and for $x=-2,f(x)=-4$, $-2>-4$, so it is also increasing for some $x<-2$, not valid.

Step4: Check third table

In the interval $(-2,1)$, when $x=-2,f(x)=-7$ and when $x=-1,f(x)=-6$, increasing. But when $x = 0,f(x)=-6$, not increasing strictly in $(-2,1)$.

Step5: Check fourth table

In the interval $(-2,1)$, when $x=-2,f(x)=7$ and when $x=-1,f(x)=1$, decreasing in part of $(-2,1)$, not valid.