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If x > 0, what is the least possible value for x + 1/x

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If x > 0, what is the least possible value for x + 1/x  [#permalink]

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New post 10 Apr 2006, 10:45
1
2
If x > 0, what is the least possible value for x + 1/x

(A) 0.5
(B) 1
(C) 1.5
(D) 2
(E) 2.5

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New post 10 Apr 2006, 10:55
It's 2. min(x + 1/x) occurs at x=1.

Correct choice is D.
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New post 10 Apr 2006, 10:58
D

Set the derivative equal to 0, and solve for x (I know we don't need to know Calculus, but sometimes it helps).

y = x + 1/x = x + x^-1

d/dx = 1 - x^-2 = 0

x^-2 = 1

x^2 = 1 or -1

Because we know x>0, then it must be 1. Which tells us that the answer is 2.
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New post 10 Apr 2006, 11:18
D is the correct answer

if x=1 => x+1/x=2, E ruled out :)
if x=0.5 => x+1/x=2.5
if x=2 => x+1/x=2.5

So, we get U shaped graph with X=1 min.
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New post 10 Apr 2006, 11:20
For some reason I have very bad habit of visualizing equations in graphs and figures ! you never know when you will be wrong. trying hard to get out of this habit :(
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New post 10 Apr 2006, 12:04
yes, its D, go with jcgoodchild explanation
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New post 11 Apr 2006, 04:17
This is a maxima minima problem. Another way of solving this is (though GMAT does not advocate the use of calculus) is:

f(x) = x + 1/x
For max x, f'(x) = 0 (i.e. the first derivative of f(x) should equal zero)
f'(x) = (d/dx)f(x) = 1 - 1/x^^2
So, if f'x) = 0 ==> 1 - 1/x^^2 = 0 ==> x = +/-1
Since x > 0, x = 1.

This is quicker than plugging in values for x.
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New post 11 Apr 2006, 06:24
Another way is using quadratic equation ;)

x+1/x = n -------- (where "n" is the minimum value we are looking for)

=> x^2-nx + 1 = 0

the descriminator = n^2-4.
For x to be real positive, 'n' can't be less than 2. So answer should be, n >= 2.
Hence 2.
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New post 11 Apr 2006, 08:24
D seems to be right.
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Re: If x > 0, what is the least possible value for x + 1/x  [#permalink]

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New post 04 Mar 2018, 06:55
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Re: If x > 0, what is the least possible value for x + 1/x &nbs [#permalink] 04 Mar 2018, 06:55
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