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 Author: questioner [ Sat Mar 31, 2012 12:48 pm ] Post subject: GMAT Functions If function f(x) = x² + 2bx + 4, then for any b, f(x) is the least when x equalsA. -b – √(b² – 4)B. -2C. 0D. -bE. b – 4(D) x² + 2bx + 4 = (x² + 2bx + b²) + 4 – b² = (x + b)² + (4 – b²).Any square is a non-negative number, so f(x) is the least, when (x + b)² = 0.x + b = 0x = -bThe correct answer is D.----------Why? I have solved for x and I got the A solution. Since it has to be the least value I've chosen the solution with "–".

Author:  Gennadiy [ Sat Mar 31, 2012 1:17 pm ]
Post subject:  Re: GMAT Functions

Quote:
Since it has to be the least value I've chosen the solution with "–".
The question does NOT ask to find the least value of x. It asks to find such value of x that f(x) will be the least.

Quote:
I have solved for x and I got the A solution.
A is a trap choice. Apparently you solved 0 = x² + 2bx + 4, which corresponds to f(x) = 0. But, why must the least value of f(x) be 0? There are no logical grounds for that.
The least value of f(x) can also be positive or negative depending on the value of b. Pick some values for b between -2 and 2. Then pick some values for b greater than 2. You'll see that the least value of f(x), which is (4 – b²), will have a different sign.

Furthermore, choice A can NOT be applied if b is between -2 and 2, while question statement asks "for any b". That is a hint that choice A is NOT a correct one.

Here is the image of general view of such parabolas that a quadratic function can be. Take a look at their bottom points that correspond to the least value for each parabola. Such point can be below the x-axis (negative), above the x-axis (positive) or lie on the x-axis (zero).