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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 [#permalink]  01 Sep 2013, 09:35
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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
[Reveal] Spoiler: OA
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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  01 Sep 2013, 10:56
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Expert's post
Revenge2013 wrote:
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 $$x=-b^2$$.

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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  01 Sep 2013, 13:08
Bunuel wrote:
Revenge2013 wrote:
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 $$x=-b^2$$.

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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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  02 Sep 2013, 05:03
Hi Bunuel,

I am still not clear on how you arrived on the following:

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.

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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  02 Sep 2013, 08:15
Bunuel wrote:
Revenge2013 wrote:
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 $$x=-b^2$$.

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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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  02 Sep 2013, 23:43
1
KUDOS
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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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  07 Sep 2013, 10:45
1
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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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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b [#permalink]  02 Nov 2014, 22:36
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Re: If f(x)=x^2/b^2 + 2x + 4, then for each non-zero value of b   [#permalink] 02 Nov 2014, 22:36
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