Is x/3 + 3/x > 2?
1) x<3
2) x>1
Determine the CRITICAL POINTS.
A critical point occurs when the inequality is UNDEFINED or when the two sides of the inequality are EQUAL.
If x=0, then 3/x is undefined.
If we test a few small values for x -- x=1, x=2, x=3 -- we'll quickly find that x/3 + 3/x = 2 when x=3.
Thus, the critical points are x=0 and x=3.
To determine where x/3 + 3/x > 2, test one value to the left and one value to the right of each critical point.
x<0:
If we plug x=-3 into x/3 + 3/x > 2, we get:
-3/3 + 3/-3 > 2
-2 > 2.
Doesn't work.
Thus, x<0 is not a valid range.
0<x<3:
If we plug x=1 into If we plug x=1 into x/3 + 3/x > 2, we get:
1/3 + 3/1 > 2
10/3 > 2.
This works
Thus, 0<x<3 is a valid range.
x>3:
If we plug x=6 into x/3 + 3/x > 2, we get:
6/3 + 3/6 > 2
5/2 > 2.
This works.
Thus, x>3 is a valid range.
Result:
x/3 + 3/x > 2 if 0<x<3 or x>3.
Question rephrased:
Is 0<x<3 or x>3?
Statement 1: x<3
If x=2, then 0<x<3.
If x=-1, then x is not in either valid range.
INSUFFICIENT.
Statement 2: x>1
If x=2, then 0<x<3.
If x=3, then x is not in either valid range.
INSUFFICIENT.
Statements combined:
Since 1<x<3, we know that x is contained with the valid range 0<x<3.
SUFFICIENT.
The correct answer is
C.
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