Which of the following inequalities has a solution set that

[tomslawsky]

Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

1) x^4 > 1

2) x^3 < 27

3) x^2 > 16

4) 2 < |x| < 5

5) 2 < 3x + 4 < 6

Im guessing E, but i'm not sure about the testing technique to use here. Please help, thank you.

[snipertrader]

So we are looking for finite values of X

A. x^4 >= 1

x >= 1 OR x <= -1 - wrong

B. x^3 <= 27

x <= 3 - x could take any negative no less than/equal to 3 - wrong

C. x^2 >= 16

x >= 4 OR x <= -4 wrong


D. 2<= |x| <= 5

2 <= x <= 5
OR
-2 >= x >= -5 - two options/line segments, wrong

Stop, Now E has to be the answer...but lets solve it to check

E. 2 <= 3x+4 <= 6

-2 <= 3x <= 2
-2/3 <= x <= 2/3 (dividing by 3)

So E it is

[RonPurewal]

So we are looking for finite values of X

A. x^4 >= 1

x >= 1 OR x <= -1 - wrong

B. x^3 <= 27

x <= 3 - x could take any negative no less than/equal to 3 - wrong

C. x^2 >= 16

x >= 4 OR x <= -4 wrong


D. 2<= |x| <= 5

2 <= x <= 5
OR
-2 >= x >= -5 - two options/line segments, wrong

Stop, Now E has to be the answer...but lets solve it to check

E. 2 <= 3x+4 <= 6

-2 <= 3x <= 2
-2/3 <= x <= 2/3 (dividing by 3)

So E it is

yes.

note that (e) is the easiest of these equations to solve (it's a routine linear inequality).
as soon as you figure out its solution set, you're done, since it's a single line segment of finite length.

[navdeep_bajwa]

IS MY LOGIC RIGHT

1) X^4 > 1
X^4 is always positive so X^4 can be written as |X|^4>1^4 OR |X|>1 which means X>1 OR X<-1 put values of X>1 or X<-1 IN THE INEQUALITY TO SEE IF IT HOLDS

2) X^3 < 27
X^3<3^3 which means X<3 put values of X<3 IN THE INEQUALITY TO SEE IF IT HOLDS

3) X^2 > 16
X^2 >4^2 => |X|^2>4^2 AS X^2 IS ALWAYS POSITIVE =>|X|>4=>X>4 OR X<-4 AGAIN
put values of X>4 AND X<-4 IN THE INEQUALITY TO SEE IF IT HOLDS

4) 2 < |X| < 5
|X|>2 => X>2 OR X<-2 AND |X|<5=> -5<X<5

5) 2 < 3X + 4 < 6
-2/3<X<2/3

[esledge]

IS MY LOGIC RIGHT
.
.
.

4) 2 < |X| < 5
|X|>2 => X>2 OR X<-2 AND |X|<5=> -5<X<5

Your logic is correct, though I think it is a bit incomplete for choice (D). Shade a number line with one color for the regions x>2 and x<-2, and another color for the region -5<x<5. The AND means the final answer is the region(s) that is(are) shaded in both colors. The final answer is -5<x-2 or 2<x<5.