GMAT Prep - Quant 20

[gusgrinberg]

For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers of a and b?

a) f(x) = x^2
b) f(x) = x+ 1
c) f(x) = x^1/2
d) f(x) = 2/x
e) f(x) = -3x

E is the correct answer.

Can you please help me solve this question? What are the concepts being touched on this particular problem?

[atharshiraz]

For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers of a and b?

a) f(x) = x^2
b) f(x) = x+ 1
c) f(x) = x^1/2
d) f(x) = 2/x
e) f(x) = -3x

E is the correct answer.

Can you please help me solve this question? What are the concepts being touched on this particular problem?

Choice (a) When f(x) = x^2
a and b are x so:
f(a=x)=x^2 (eq 1)
and
f(b=x) = x^2 (eq 2)
and
f((a=x) + (b=x)) = f(x +x) = (2x)^2= 4x^2 (LHS)
so using (eq 1) and (eq 2)
f(a)+f(b) = x^2 + x^2 = 2x^2 (RHS)

So LHS != RHS (the left and right hand side differ)

We do the same for all the other choices (feel free to ask if you don't know how to do it for any of them) or if there is any lack of clarity either I or someone else would be more than happy to clarify.

Coming to choice (e) When f(x) = -3x
a and b are x so:
f(a)=-3x (eq 1)
and
f(b) = -3x (eq 2)
and
f(a + b) = f(x+x) = -3(2x)= -6x (LHS)
so using (eq 1) and (eq 2)
f(a)+f(b) = -3x -3x = -6x (RHS)

So LHS == RHS and e is the answer.

[gusgrinberg]

Gotcha! Thanks much!

[RonPurewal]

search the forum, people

post9694.html?sid=16be265d66bf2c6bd5d0028e2ef57e6c#p9694

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to the original poster:
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