Problem Solving

If x is +ve , which of the following could be the correct ordering of 1/x , 2x and x^2
I. x^2 <2x <1/x
II x^2 <1/x <2x
III 2x <x^2 <1/x

a) None
b) I only
c) III only
d) I and II only
e) I,II and III

OA D, I am able to get I but not II

If x = .75 then you get 2.

If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x²?

I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

a. None
b. I
c. III
d. I and II
e. I, II, and III

Determine the CRITICAL POINTS by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points are where two of the expressions are equal.
Thus, to the right and left of each critical point, the value of one expression is GREATER than the value of another.

To determine which answer choices are possible, PLUG IN ONE VALUE TO THE LEFT AND RIGHT OF EACH CRITICAL POINT.

x < 5/7:
If x=1/2, then:
1/x = 2.
x² = 1/4.
2x = 1.
Since x² < 2x < 1/x, we know that I could be true.
Eliminate A and C.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate B.

In III, the largest value listed is 1/x.
For 1/x to be the largest value, x would have to be a fraction.
Having tried a fraction on each side of the critical point of 5/7, we know that there is no way that III could be true.

The correct answer is D.

Great explanation.

Thanks!!