In a question like this when a constraint is given to you such as x must be in between zero and -1 therefore x must be negative greatr than -1 but less then zero, in other wirds -1/2 or -1/3 or similar negatives values

I hope this helps

I follow that, but let's use x = -1/2 for examples. x^3 = -1/8 and 1/x^3 = -8. Both negative, but X^3 has a smaller absolute value while 1/x^3 is more negative. OE implies that "smaller" means more negative. Is that always the case in GRE Quant?

Suppose that x is \(-\frac{1}{2}\)

A) \(\frac{1}{x^3}=\frac{1}{-\frac{1}{8}}=-8\)

B) \(\frac{1}{x^2}=\frac{1}{\frac{1}{4}}=4\)

C) \(-\frac{1}{2}\)

D) \(\frac{1}{4}\)

E) \(-\frac{1}{8}\)

On the number line, we have

-8 -------------------- - 1/2------------ - 1/8 -------- 0 ------------ 1/4 ---------------------------- 4

Clea<rly A is the answer which is the smallest value

I hope this helps

Hey Carcass, thanks for taking the time to answer my question. I understand that in this problem, smallest means most negative (e.g., \(−8\) is smaller than \(−\frac{1}{8}\)). What I’m trying to understand is whether this is always how “smallest” is defined on the GRE, or if it could ever mean smallest in the absolute value sense. If I interpreted smallest in the absolute value sense, I would be inclined to pick E.

Unless stated - which i never saw in a question - smallest is the lowest value. Regardless absolute value.

In this case is -8 not \(|-8|\) which is 8