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Revenge2013
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Bunuel
Revenge2013
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

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.

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!
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Bunuel
Revenge2013
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

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.


Bunuel,

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

so the OA is D
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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.
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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
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Revenge2013
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

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.
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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.
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