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is x^2 + x + 1 minimum?

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by sanju09 » Fri Oct 15, 2010 2:04 am
For what value of x is x^2 + x + 1 minimum?
(A) 1
(B) ½
(C) 0
(D) -½
(E) -1


[spoiler]Made up[/spoiler]
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by shovan85 » Fri Oct 15, 2010 2:54 am
Is it a GMAT Q? I am not sure but it requires derivative of the eqn. Or one can try the values substituting into the eqn.

Derivative of eqn = 0 then the solution to this derived eqn is minimum for the main eqn.

d(x^2 + x + 1)/dx = 2x+1 = 0
=> x = -1/2

IMO D
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by neerajkumar1_1 » Fri Oct 15, 2010 3:02 am
well the derivation method was absolutely perfect...
but for those who are not familiar with it...
one could just plug in the answers...

with x = 1 value of exp = 3
with x = 1/2 value of exp = >1
with x = 0 value of exp = 1
with x = -1/2 value of exp = <1
with x = -1 value of exp = 1

min at -1/2

Pick D
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by gmatmachoman » Fri Oct 15, 2010 3:29 am
IMO D
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by sanju09 » Fri Oct 15, 2010 5:44 am
I don't think that Differential Calculus is the only resort for such problems, otherwise I would have thought twice before figuring it out for here. Getting D by plugging is not a bad idea on GMAT test either, but just because x^2 + x + 1 is minimum at D among the available answers, we can't otherwise say that this is the only value at which the function will be minimum, and hence, we could have probably not answered the stem. Even if plugging-in works, some GMAT questions are not worded so. This question was ambiguous had it been designed to be cracked by the pick-n-plug policy, the wording clearly anticipates us to zero in to that value of x without breaking GMAT barriers.

See

x^2 + x + 1

= x^2 + 2 × ½ × x + ¼ + ¾

= (x + ½) ^2 + ¾

To minimize this, we need to minimize (x + ½) ^2, which being a perfect square, cannot be less than zero. Hence, minimum (x + ½) ^2 = 0 and this is at [spoiler]x = - ½


D[/spoiler]

And if the question is changed to...

What is the minimum value of the expression x^2 + x + 1?

The same procedure would have given us [spoiler]¾[/spoiler] as the right answer?
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by neerajkumar1_1 » Fri Oct 15, 2010 6:55 am
sanju09 wrote:I don't think that Differential Calculus is the only resort for such problems, otherwise I would have thought twice before figuring it out for here. Getting D by plugging is not a bad idea on GMAT test either, but just because x^2 + x + 1 is minimum at D among the available answers, we can't otherwise say that this is the only value at which the function will be minimum, and hence, we could have probably not answered the stem. Even if plugging-in works, some GMAT questions are not worded so. This question was ambiguous had it been designed to be cracked by the pick-n-plug policy, the wording clearly anticipates us to zero in to that value of x without breaking GMAT barriers.

See

x^2 + x + 1

= x^2 + 2 × ½ × x + ¼ + ¾

= (x + ½) ^2 + ¾

To minimize this, we need to minimize (x + ½) ^2, which being a perfect square, cannot be less than zero. Hence, minimum (x + ½) ^2 = 0 and this is at [spoiler]x = - ½


D[/spoiler]

And if the question is changed to...

What is the minimum value of the expression x^2 + x + 1?

The same procedure would have given us [spoiler]¾[/spoiler] as the right answer?
nice work..
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by fskilnik@GMATH » Tue Oct 19, 2010 4:00 am
sanju09 wrote:I don't think that Differential Calculus is the only resort for such problems, otherwise I would have thought twice before figuring it out for here. Getting D by plugging is not a bad idea on GMAT test either, but just because x^2 + x + 1 is minimum at D among the available answers, we can't otherwise say that this is the only value at which the function will be minimum, and hence, we could have probably not answered the stem. Even if plugging-in works, some GMAT questions are not worded so.
Well said, Sanju!

Great solution, by the way. Let me show another one, perhaps even more general... I hope you (all) like it!

Regards,
Fabio.

Alternative solution: let y = x^2+x +1 then points (x,y) on the rectangular system xOy are on a parabola. The minimum value of x is attained at its vertex (where y is also minimum, by the way), therefore it is enough to calculate -b/(2a) = -1/2.

Suggestion: to remember the x-vertex formula, imagine that the equation x^2+x+1 = 0 has discriminant non-negative, then we may say x1 and x2 are its (real) roots, to affirm that the x-vertex is in their middle point, that is, (x1+x2)/2. From the fact that x1+x2 = -b/a (part of the "sum/product" technique to find the roots) then the x-vertex is -b/(2a). When the equation has discriminant negative (this is the case in this problem), the x-vertex formula is still valid. The reason is beautiful .... think about y = ax^2+bx+c and the importance of (changing) the "c" coefficient (leaving a and b "freezed").
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by fskilnik@GMATH » Tue Oct 19, 2010 4:04 am
sanju09 wrote:And if the question is changed to...

What is the minimum value of the expression x^2 + x + 1?

The same procedure would have given us [spoiler]¾[/spoiler] as the right answer?
Sure, by "your" method and by "mine"! :)

(That´s why I do not suggest students to know by heart the y-vertex formula; it is enough to know the x-vertex formula --or better, deduce it the way I told you -- and then substitute x in the y = ax^2+bx+c expression by the x-vertex to find the y-vertex value...)
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by Stuart@KaplanGMAT » Tue Oct 19, 2010 9:46 am
sanju09 wrote:For what value of x is x^2 + x + 1 minimum?
(A) 1
(B) ½
(C) 0
(D) -½
(E) -1


[spoiler]Made up[/spoiler]
While not worded exactly like a GMAT question (at the very least, GMAT questions are grammatically correct), a similar question could appear on the test.

We can avoid calculus and coordinate geometry by applying the most useful of all GMAT tools - logic/common sense.

"1" is a constant, so is irrelevant to the question. We can simply ask:

"What value of x minimizes the value of x^2 + x?"

Peeking at the choices (something we always want to do on test day), we can quickly eliminate A and B, since x=0 makes the expression equal to 0 and a positive value for x will make the expression greater than 0.

So, we only need to think about or plug in D and E.

When plugging in, always start with the easiest choices - let's try E.

if x = -1, then x^2 + x = 1 - 1 = 0. Awesome!

Why awesome? Because E provides the same value as C. Since there can't be two correct answers to a question, they both must be incorrect. Choose D, no further work required.
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