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Re: A function is defined for all positive numbers x as f(x) = ax^(1/2) + [#permalink]
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Bunuel wrote:
A function is defined for all positive numbers x as \(f(x) = a\sqrt{x} + b\). What is the value of f(3), if f(4) – f(1) = 2 and f(4) + f(1) = 10?


(A) 1

(B) 2

(C) \(2 \sqrt{3}\)

(D) \(2 \sqrt{3} + 2\)

(E) \(2 \sqrt{3} – 2\)


f(4) – f(1) = 2 ==> \(a\sqrt{4} + b\) - \(a\sqrt{1} - b\) ==> 2a - a =2 ==> a=2

f(4) + f(1) = 10 ==> \(a\sqrt{4} + b\) + \(a\sqrt{1} + b\) ==> 6+2b =10 ==> b=2

Hence D

f(3) = \(2 \sqrt{3} + 2\)
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Re: A function is defined for all positive numbers x as f(x) = ax^(1/2) + [#permalink]
How does \(a\sqrt{4} + b\) + \(a\sqrt{1} + b\) ==> 6+2b =10??

NandishSS wrote:
Bunuel wrote:
A function is defined for all positive numbers x as \(f(x) = a\sqrt{x} + b\). What is the value of f(3), if f(4) – f(1) = 2 and f(4) + f(1) = 10?


(A) 1

(B) 2

(C) \(2 \sqrt{3}\)

(D) \(2 \sqrt{3} + 2\)

(E) \(2 \sqrt{3} – 2\)


f(4) – f(1) = 2 ==> \(a\sqrt{4} + b\) - \(a\sqrt{1} - b\) ==> 2a - a =2 ==> a=2

f(4) + f(1) = 10 ==> \(a\sqrt{4} + b\) + \(a\sqrt{1} + b\) ==> 6+2b =10 ==> b=2

Hence D

f(3) = \(2 \sqrt{3} + 2\)
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Re: A function is defined for all positive numbers x as f(x) = ax^(1/2) + [#permalink]
Bunuel wrote:
A function is defined for all positive numbers x as \(f(x) = a\sqrt{x} + b\). What is the value of f(3), if f(4) – f(1) = 2 and f(4) + f(1) = 10?


(A) 1

(B) 2

(C) \(2 \sqrt{3}\)

(D) \(2 \sqrt{3} + 2\)

(E) \(2 \sqrt{3} – 2\)


Bunuel,

Why doesn't it work algebraically to add the two functions together?

f(4)-f(1)=2
+
f(4)+f(1)=10
-------------------

f(1) cancels out, leaving (2)*f(4) = 12, which reduces to f(4) = 6. When subbing 6 into the second equation, then f(1) = 4.

So I ended up with f(3) = 6([square_root]3) +4

Did I fall into some sort of trap by trying to solve it this way? I need to understand why this doesn't work so I don't do it again in the future.
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Re: A function is defined for all positive numbers x as f(x) = ax^(1/2) + [#permalink]
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