Problem Solving

Wich of the follow inqualities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A. X^4 > 1
B. x^3 < 27
C x^2 > 16
D. 2 < | x | < 5
E. 2 < 3x + 4 < 6

The < and > are equal to, Don't know how to do that symbol :$

Eliminate the first three options quickly:
A. x^4 > 1 is valid for any number greater than 1 or smaller than -1, so we have an infinity of solutions
B. x^3 < 27 is valid for any number smaller than 3 (3^3 = 27), including all negative numbers. So this one also has an infinity of solutions
C. x^2>=16 also has an infinity of solutions, since all numbers greater than 4 or smaller than -4 are consistent with this inequality.

My hunch is that you got confused by the last two options.
D. 2 <= |x| <= 5 has two cases, depending on the sign of x:
a. x is negative makes |x| = -x, making the inequality 2 <= -x <= 5. Multiply by -1 and you get that -2 >= x >= -5, or that x is between -5 and -2. This is a finite segment.
b. x is positive, with |x| = x, giving us that x is between 2 and 5. This is also a finite segment.
But if you put together the two cases, you get that the solution is actually two finite segments, so it's not what you're looking for.

I would advise just stopping here and picking E, since it's the only solution left. But we can demonstrate that:
2 <= 3x + 4 <=6
-2 <= 3x <= 2
-2/3 <= x < 2/3 ----- finite segment between -2/3 and 2/3.

A + B + C gives a parabola?

Yeah, but that doesn't even matter, since they have an infinite number of solutions, which makes your segment infinite as well.... Let's not forget we're looking for finite segments here.

IMO E

A, B, C gives us only one limit.

D gives us two segments


E gives us one finite segment.