Question

which table shows a function that is increasing only over the interval ((-2,1)), and constant elsewhere? (the tables are presented with (x) and (f(x)) columns. the first table: (x) values -3, -2, -1, 0, 1, 2; (f(x)) values -2, -4, -1, 1, 4, 3. the second table: (x) values -3, -2, -1; (f(x)) values -6, -3, -1. the third table: (x) values -3, -2, -1, 0, 1, 2; (f(x)) values -3, -5, -7, -6, 1, -1. the fourth table: (x) values -3, -2, -1, 0, 1, 2; (f(x)) values 5, 7, 1, 0, -4, -2.)

Explanation:

Step1: Define increasing/decreasing

A function increases if $x_1f(x_2)$. We check interval $(-2,1)$: $x=-1,0,1$.

Step2: Check Table 1 (top-right)

For $x=-1,0,1$: $f(-1)=-1$, $f(0)=1$, $f(1)=4$.
$-1<0<1$ and $-1<1<4$: increasing on $(-2,1)$.
Check other $x$: $x=-3,-2$: $f(-3)=-2$, $f(-2)=-4$; $-3<-2$ but $-2>-4$ (decreasing). $x=1,2$: $f(1)=4$, $f(2)=3$; $1<2$ but $4>3$ (decreasing).

Step3: Verify other tables

• Top-left: Only $x=-3,-2,-1$: no $(-2,1)$ coverage.
• Middle: $x=-1,0,1$: $f(-1)=-7$, $f(0)=-6$, $f(1)=1$ (increasing), but $x=-3,-2$: $f(-3)=-3$, $f(-2)=-5$ (decreasing); $x=1,2$: $f(1)=1$, $f(2)=-1$ (decreasing) – but wait, no, check full trend: no, wait, the top-right is the only one where ONLY $(-2,1)$ is increasing.
• Bottom: $x=-1,0,1$: $f(-1)=1$, $f(0)=0$, $f(1)=-4$ (decreasing), so invalid.

Answer:

The table with columns:

$x$	$f(x)$
-2	-4
-1	-1
0	1
1	4
2	3