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ps - function

by ccassel » Thu Mar 31, 2011 12:17 pm
How would you explain the answer to this question?

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

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

cheers,
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by MAAJ » Thu Mar 31, 2011 12:34 pm
Correct Answer [spoiler](E)[/spoiler]

(a) f(x)=x^2
f(a+b) = (a+b)²
f(a)= a²
f(b)= b²
(a+b)² = a² + b²? NO

(b) f(x)=x+1
f(a+b) = (a+b+1)
f(a) = a+1
f(b) = b+1
(a+b+1)= a+1 + b+1? NO

(c) f(x)=sqrt(x)
f(a+b) = sqrt(a+b)
f(a) = sqrt(a)
f(b) = sqrt(b)
sqrt(a+b) = sqrt(a) + sqrt(b)? NO

(d) f(x)=2/x
f(a+b) = 2/(a+b)
f(a) = (2/a)
f(b) = (2/b)
2/(a+b) = (2/a) + (2/b) ? NO

(e) f(x)=-3x
f(a+b) = -3(a+b) = -3a - 3b
f(a) = -3a
f(b) = -3b
-3a - 3b = -3a -3b? YES

Do you have to do all this? No, I would have started by option (b) or (e) 'cause they are more likely to be the correct answer.
Last edited by MAAJ on Thu Mar 31, 2011 12:38 pm, edited 1 time in total.
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by emalby » Thu Mar 31, 2011 12:36 pm
(a) f(x)=x^2 => means you have to substitute x with (a+b) or (a) or (b) so in this case f(a+b) = (a+b)^2, while f(a) = a^2 and f(b) = b^2

So is (a+b)^2 = a^2 + b^2 ? Well, (a+b)^2 = a^2 + b^2 + 2ab which is different from a^2 + b^2 because it contains the factor 2ab.

Therefore we eliminate (a).

Suggest you exercise with the others. The correct one is [spoiler](e)[/spoiler].
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by ccassel » Thu Mar 31, 2011 1:06 pm
Thanks for clarifying the meaning of the question.

Solving it is alot easier when you understand the language!

Cheers,
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by sensei_mike » Thu Mar 31, 2011 1:06 pm
What I would do for this q is plug in some numbers to make life easier. So pretend a=2 and b=3, plug into each answer choice and see which works.
E: f(2+3=5)= -15 and f(2)+f(3)= -6 + (-9)= -15

So they equal each other and thus is the correct answer.
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by force5 » Thu Mar 31, 2011 1:16 pm
Yes its definitely E.
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