If x 0, what is the smallest possible value of 2x+12x ?

Question

If x > 0, what is the smallest possible value of  2x+\frac{1}{2x} ?

A. 0.5
B. 1
C. 1.5
D. 2
E 2.5

How to approach such questions?

nick1816 · Accepted Answer

If you're uncomfortable with AM-GM inequality, you can use the following approach (both are one and the same thing tho)

2x + \frac{1}{2x} -2 +2

(\sqrt{2x })^2+ (\sqrt{\frac{1}{2x}})^2 -2\sqrt{2x*\frac{1}{2x}} +2

^2 +2

Square of any number is non-negative.

hence  2x+\frac{1}{2x}  is always greater than or equal to 2.

You can generalize it too. Sum of any positive number and its reciprocal is always greater than or equal to 2.

nick1816 · Answer

/2 ≥ \sqrt{2x*\frac{1}{2x}}

2x+\frac{1}{2x} ≥ 2

Kritisood · Answer

nick1816  could you please elaborate on how you approach such questions?

Kinshook · Answer

Asked: If x > 0, what is the smallest possible value of  2x+\frac{1}{2x} ?

SInce AM>GM

(2x + 1/2x)/2 >= \sqrt{2x * 1/2x } = 2

IMO D

hasanloubani · Answer

Bunuel  can you please solve this with explanation?

Bunuel · Answer

Notice that  2x * \frac{1}{2x} = 1 . For a given product, the sum of the positive factors is minimized when the factors are equal.

So, given that  2x * \frac{1}{2x} = 1  (product),  2x+\frac{1}{2x}  (the sum) is minimized when  2x = \frac{1}{2x} . Solving gives x = 1/2.

2x+\frac{1}{2x}=2 .

Or using derivatives ( don't need to study this for GMAT ):   \frac{d}{dx}(2x++\frac{1}{2x})=\frac{d}{dx}(2x+(2x)^{(-1)}) = 2 - \frac{1}{2x^2} = 0  --> x = 1/2.

2x+\frac{1}{2x}=2 .

Answer: D.

hasanloubani · Answer

The first way worked very well. Thank you

David nguyen · Answer

TBH, There are too many formulas to remember. And I do not remember any formulas that I can apply on this question. So I decided to solve this by using logic. Look at 2x + \frac{1}{2x} for a second, I see that: If x = \frac{1}{2 } The value of (2x + \frac{1}{2x} ) = 1 + 1 = 2 Now, Look again: It does not matter what the value of \frac{ 1}{2x} is, as long as X > 1, the value of 2x always >2 => 2x + \frac{1}{x} >2 Hence, possible range of x to get to 2 => 0 2 If X < \frac{1}{2.5} or X > \frac{2}{3 } Hence, possible range of x to get to 2 is now narrower => \frac{1}{(2.5)} < x < \frac{2}{3} One can conclude that the further x moves away from 1/2, the greater than 2 the value of 2x + \frac{1}{2x} is. Therefore, the smallest value of 2x+ \frac{1}{2x} must be 2 or slightly close to 2. Checking the answer choice, D =2, C and E are too far away from 2. D Must be the answer.

Avinash514043 · Answer

Derivative is 2-(0.5/x^2) =0 
The value of x here will yield the min. Value of x for which the function will become minimum.
 X=0.5

Now put the value of X in main function that will be 2.
Option D.

Posted from my mobile device

Kritisood · Answer

Hi  Bunuel  i understood the part highlighted in green but what's the significance of the part in yellow?

Bunuel · Answer

We know the product of two numbers ( 2x * \frac{1}{2x} = 1 ) and we want to minimize their sum. The point is that the product is fixed, it's some number, not dependent on x, so we can apply the property discussed.

PallabiKundu · Answer

Can we answer the question in this way?

since x>0 , we can take x=1/2. 
Now, substituting the value of x=1/2 in the equation 2x+1/2x we get the value of the equation as 2, which is the required answer D.

Bunuel  can u pls suggest if I am right in the explanation?

Bunuel · Answer

No. With that approach, how would you know that for other values of x the expression will not be less than 2?

NatanBricks222 · Answer

Omg, I'm too silly to solve this

SohamS · Answer

if x = 1/2 = 0.5, then 2x + 1/2x = (2 * 1/2 )+ 1/ (2 * 1/2) = 1 + 1 = 2. So this will be the lowest value of the equation right?

Fdambro294 · Answer

Let the expression = q

2x + (1 / 2x) = q

— multiply both sides of the equation by 2x——

4(x)^2 + 1 = 2q(x)

4(x)^2 - 2q(x) + 1 = 0

For there to be a Root of the quadratic equation, the following must be true:

(b)^2 - 4ac >/= 0

plugging in the coefficients:

(-2q)^2 - 4(4)(1) >/= 0

4(q)^2 >/= 16

—-taking the square root of each side, realizing they each answer provided must be non-negative so only worrying about the positive values that Q can take———

2(q) >/= 4

q >/= 2

Plug in q = 2 as proof. X will be equal to 1/2.

(D)

The minimum possible value of the expression = 2

Posted from my mobile device

shruti_tapadia · Answer

KarishmaB  can you please explain this

KarishmaB · Answer

The minimum value of a + 1/a where a is a positive number is 2. The relation is useful enough to be kept in mind.

Why is it so? Because when product of two positive numbers is constant, the sum is minimum when the numbers are equal.

Product of 2x and 1/2x is constant. 
 2x * \frac{1}{2x} = 1

So sum is minimum when  2x = \frac{1}{2x} 
i.e. when x = 1/2

So minimum value of  2x+\frac{1}{2x} = 2*\frac{1}{2} + \frac{1}{2*\frac{1}{2}} = 2

Answer (D)

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x > 0, what is the smallest possible value of 2x+12x ?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions