Question

which table shows a function that is decreasing only over the interval (-1, ∞)?

x	f(x)		x	f(x)		x	f(x)
-3	-1		-3	-3		-3	-4
-2	-3		-2	-5		-2	-3
-1	-5		-1	-7		-1	-1
0	-2		0	-6		0	2
1	-1		1	1		1	1
2	2		2	-1		2	-6

x	f(x)
-3	-5
-2	-1
-1	1
0	0
1	-4
2	-8

Explanation:

Step1: Define decreasing function

A function is decreasing over an interval if for any \(x_1 < x_2\) in the interval, \(f(x_1) > f(x_2)\). We need a function that is decreasing only over \((-1, \infty)\), meaning it is not decreasing for \(x \leq -1\).

Step2: Check Table 1 (top-left)

• \(x \leq -1\): \(x=-3\to f=-1\), \(x=-2\to f=-3\), \(x=-1\to f=-5\) (decreasing here, violates "only over \((-1,\infty)\)").

Step3: Check Table 2 (top-middle)

• \(x \leq -1\): \(x=-3\to f=-3\), \(x=-2\to f=-5\), \(x=-1\to f=-7\) (decreasing here, violates).

Step4: Check Table 3 (top-right)

• \(x \leq -1\): \(x=-3\to f=-4\), \(x=-2\to f=-3\), \(x=-1\to f=-1\) (increasing here, good so far).
• \(x > -1\): \(x=0\to f=2\), \(x=1\to f=1\), \(x=2\to f=-6\) (decreasing here, but wait: \(x=0\) to \(x=1\): \(2>1\) (decreasing), \(x=1\) to \(2\): \(1>-6\) (decreasing). But wait: Is it decreasing only here? Yes for \(x\leq-1\) it's increasing, but wait—wait Table 4:

Step5: Check Table 4 (bottom-right)

• \(x \leq -1\): \(x=-3\to f=-5\), \(x=-2\to f=-1\), \(x=-1\to f=1\) (increasing here, good).
• \(x > -1\): \(x=0\to f=0\), \(x=1\to f=-4\), \(x=2\to f=-8\): \(1>0> -4> -8\) (decreasing here). And it is not decreasing for \(x\leq-1\) (it's increasing there). This fits "decreasing only over \((-1,\infty)\)".

(Note: Table 3 has \(x=0\to2\) which is higher than \(x=-1\to-1\), but Table 4's \(x\leq-1\) is strictly increasing, and \(x>-1\) is strictly decreasing—no other intervals where it's decreasing. So Table 4 is correct.)

Answer:

The fourth table (bottom-right table)