If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b

Question

If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b the function of f(x) is the least when x equals

A.  -1 -\sqrt{(1-4b)} 
B. -2
C. 0
D. -b^2
E. b-4

Bunuel · Accepted Answer

Complete the square.

f(x)=\frac{x^2}{b^2} + 2x + 4  --> add b^2-b^2 -->  f(x)=\frac{x^2}{b^2} + 2x + b^2 +4 -b^2  -->  f(x)=(\frac{x}{b})^2+ 2x + b^2 +4 -b^2  -->  f(x)=(\frac{x}{b}+b)^2+(4 -b^2)=(non-negative)+(constant) .

The least value of f(x) when  (\frac{x}{b}+b)^2=0 , so when  \frac{x}{b}+b=0  or when  x=-b^2 .

Answer: D.

IanStewart · Answer

I suppose you could solve by looking at the answer choices. If the answer is going to be correct for absolutely every value of b, it certainly must be correct when b=1. So we can let b=1 and just plug each answer choice into

x^2 + 2x + 4

to see which gives us the least value of this function. Notice answer A is completely undefined (we get a negative under the root), so cannot possibly be right. If we plug in -2 or 0, the value of the function is 4 in both cases. If we plug in answer D, which is -b^2 = -1, the function's value is 3, which is the smallest value so far. Finally if we plug in answer E, which would be equal to b - 4 = 1 - 4 = -3, the function's value is 6. So the correct answer is -b^2.

But if the GMAT were to ask a question like this, there would always be a way to correctly solve it 'from start to finish', without looking at the answer choices. And I don't see a way to do that here using normal GMAT math. I've never once needed to 'complete the square' in any official GMAT question I've ever solved, nor have I ever needed the quadratic formula, or needed to know how to find the minimum value of a general parabola. I've solved probably close to 10,000 official questions by now, so those techniques just aren't ever required on the test. This question is one that would normally be answered using calculus anyway (and is quite easy if you know calculus), so it's a question I'd bet GMAC would consider unfair, because people with certain educational backgrounds would have a big advantage answering it, and the GMAT is not supposed to be a test of whether you've ever taken a calculus class. It's a test of how well you can reason using only the most elementary facts of mathematics.

So unless I'm not seeing a solution here that uses normal GMAT math, there's really no reason to worry about this question.

Revenge2013 · Answer

Bunuel,

Thanks for the quick reply as always!

I still fail to see the (incomplete) square that should have triggered something in me to come up with b^2-b^2. Was the trigger the x^2/b^2?
The rest is clear.

Many thanks!

maaadhu · Answer

Bunuel,

Can we say the quadratic equation ax^2 + bx + c reaches minimum for -b/2a

so the OA is D

nitin6305 · Answer

In this question, we can see that the coefficient of x^2 is always positive, therefore the equation is a parabola facing upwards in a coordinate plane. In a parabola facing upwards there is only minima (no maxima) which is equal to (-coeff of x/2coeff of x^2), in this case -b^2. Hence the answer, D.

Some basic knowledge about coordinate geometry makes such questions cake walk.
The GMAT Club Math Book deals with such basics appropriately.

Hope it helps.

Some basic knowledge about coordinate geometry makes such questions cake walk. 
The GMAT Club Math Book deals with such basics appropriately.

Hope it helps.

ramannanda9 · Answer

D
Derivatives do the trick here as well

f(x)=x^2/b^2+2x+4
=> 2x/b^2+2=0 => x=-b^2 which gives a minimum value of -b^2+4

balaji4799 · Answer

This question can also be solved by considering the derivative of f(x).
The theory states that the function can achieve its maximum or minimum value at x when d(f(x)/dx= 0 (basically slope of the curve is 0, parallel to x axis) 
Now to find out whether the function gets max or min we need to find out double derivative of f(x) 
if the double derivative of f(x) at x which is calculated above from single derivative is <0 then the function attains maximum value
otherwise the function attains minimum value.

adewale223 · Answer

Since a is positive in the equation, parabola is facing up. So value of x at the turning point is x = -b/2a

-2 /1 x b^2 /2 = -b^2

Adewale Fasipe GMAT, GRE quant instructor from Lagos, Nigeria.

luisdicampo · Answer

Deconstructing the Question  
We have  f(x)=\frac{x^2}{b^2}+2x+4  with  b
e 0 .
This is a parabola opening upward because  \frac{1}{b^2}>0  for any nonzero  b .
So the minimum occurs at the vertex  x=-\frac{B}{2A}  for  Ax^2+Bx+C .

Step-by-step  
Identify coefficients:
 A=\frac{1}{b^2} 
 B=2

Vertex:
 x=-\frac{B}{2A}=-\frac{2}{2\cdot (1/b^2)}

Simplify:
 x=-\frac{2}{2/b^2}=-b^2

Answer: D

pappal · Answer

best and minimum time consuming method is using quadratic equation minimum value formula
i.e. the upward facing quadratic function attains a minima at x=-(coefficient of x)/2*coefficient of x^2
so x=-2/(2/b^2)=-b^2
D

If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b

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If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b

Prep Toolkit

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards