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Intern  Joined: 15 Aug 2014
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If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 55% (01:31) correct 45% (02:06) wrong based on 243 sessions

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If function f(x) = x^2 + 2bx + 4, then for any b, f(x) is the least when x equals

A. -b – √(b² – 4)
B. -2
C. 0
D. -b
E. b – 4

Source: Scoregetter Gmat Test

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Originally posted by Nipungautam23 on 01 Nov 2014, 17:36.
Last edited by Bunuel on 02 Nov 2014, 07:15, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe  [#permalink]

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Nipungautam23 wrote:
If function f(x) = x^2 + 2bx + 4, then for any b, f(x) is the least when x equals

A. -b – √(b² – 4)
B. -2
C. 0
D. -b
E. b – 4

Source: Scoregetter Gmat Test

f(x) is a quadratic in x. We know the minimum value of ax^2 + bx + c is at x = -b/2a (draw the graph if you are not sure how we arrive at this - it's good to remember this though).

So minimum value of f(x) is at x = -2b/2*1 = -b

Or

Put b = 0. f(x) = x^2 + 4. This will be minimum when x^2 = 0 since x^2 can anyway not be negative. So x = 0. Two options give 0: (C) and (D)
Now notice that when b is a large negative value such as -100, the minimum value of f(x) will not be at x = 0.
For example, f(x) = x^2 - 200x + 4
If x = 0, f(x) = 4 but if x = 1, f(x) = - 195

So answer must be option (D).
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Math Expert V
Joined: 02 Sep 2009
Posts: 58378
Re: If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe  [#permalink]

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Nipungautam23 wrote:
If function f(x) = x^2 + 2bx + 4, then for any b, f(x) is the least when x equals

A. -b – √(b² – 4)
B. -2
C. 0
D. -b
E. b – 4

Source: Scoregetter Gmat Test

$$x^2 + 2bx + 4 = (x^2+ 2bx + b^2) + 4 - b^2 = (x + b)^2 + (4 - b^2)=(nonnegative \ value) + (constant)$$. To minimize this expression we should minimize (x + b)^2. The least value of it is 0,, which is obtained when x = -b.

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Re: If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe  [#permalink]

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Hi Bunuel !!!!!

We can use differentiation approach too....I know this is out of scope but still we are bothered about getting right answer in least possible time....put df(x)/d(x) = 0.....d(x^2 + 2bx + 4)/d(x)=0....i.e. 2x + 2b = 0.......x=-b
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe  [#permalink]

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Hi All,

This question can be solved by TESTing VALUES.

We're given the function f(X) = X^2 + 2BX + 4. We're told for ANY value of B, f(X) is LEAST when X = what value.

This means that we can choose ANY value we want for B and the answer will stay the SAME. Looking at the 5 answer choices, I don't want to use B=0 or B=2 (since those two possibilities would create duplicate answers); so I'll use...

B = 1

f(X) = X^2 + 2X + 4

Which of the following answers gives us the LEAST result when X = ......

A: X = -1 --> -1 - \sqrt{-3} = nonsenscial answer. Eliminate A.

B: X = -2 --> 4 - 4 + 4 = 4

C: X = 0 --> 0 + 0 + 4 = 4

D: X = -1 --> 1 - 2 + 4 = 3

E: X = -3 --> 9 - 6 + 4 = 7

GMAT assassins aren't born, they're made,
Rich
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Re: If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe  [#permalink]

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Hi Bunuel !!!!!

We can use differentiation approach too....I know this is out of scope but still we are bothered about getting right answer in least possible time....put df(x)/d(x) = 0.....d(x^2 + 2bx + 4)/d(x)=0....i.e. 2x + 2b = 0.......x=-b

SEEMS TO BE THE BEST APPROACH. Re: If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least whe   [#permalink] 25 Mar 2019, 21:37
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