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# Functions Question

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Senior Manager
Joined: 05 Oct 2008
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08 Oct 2008, 11:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi, can someone help with understanding the following question. I am not familiar with functions and the way they work. Also, any advice on where I could get material to understand this topic.

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

f(x) = f(x square)
f(x) = x+1
f(x) = root x
f(x) = 2/x
f(x) = -3x
Senior Manager
Joined: 04 Jan 2006
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08 Oct 2008, 12:11
2
KUDOS
study wrote:
Hi, can someone help with understanding the following question. I am not familiar with functions and the way they work. Also, any advice on where I could get material to understand this topic.

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

f(x) = f(x square)
f(x) = x+1
f(x) = root x
f(x) = 2/x
f(x) = -3x

A) Since; f(x) = f($$x^2$$)
f(a) = $$a^2$$
f(b) = $$b^2$$
f(a) + f(b) = $$a^2 + b^2$$ ---------- (1)
and
f(a+b) = $$(a+b)^2$$ ---------- (2)

(1) is not equal to (2). A is incorrect.

B) Since; f(x) = x + 1
f(1) = 1 + 1 = 2
f(x) + f(1) = (x+1) + 2 = x + 3 ---------- (1)
and
f(x+1) = [x+1] + 1 = x + 2 ---------- (2)

(1) is not equal to (2). B is incorrect.

E) f(x) = -3x
f(a) = -3a
f(b) = -3b
f(a) + f(b) = -3a - 3b = -3(a+b) ---------- (1)
and
f(a+b) = -3(a+b) ---------- (2)

(1) = (2), Choice E is the correct answer.
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08 Oct 2008, 12:32
2
KUDOS
I will attack this problem by back-solving.

First, let a = 2 and b = 3. this makes a+b = 5. Substitute 5 for x, which is a+b, and substitute 2 for f(a) and 3 for f(b).

Now, go through each answer choices by plugging in these values for a, b, and a+b for each function:

a) 5^2 = 2^2 + 3^3 ? 25 = 4+9 ? Nope
b) 5+1 = (2+1) +(3+1)? 6 = 3+4 ? Nope
c) square root (5) = square root (2) + square root (3)? ~2.24 = ~1.41 + 1.73? Nope
d) 2/5 = 2/2 + 2/3 ? 2/5 = 1+ 2/3? Nope
e)-3(5) = -3(2) + -3(3)? -15 = -6 + -9? Yes

Senior Manager
Joined: 05 Oct 2008
Posts: 272
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08 Oct 2008, 12:37
Thanks, Devilmirror & Brandonslee for the fantastic explanation...

I have given you both +1
Re: Functions Question   [#permalink] 08 Oct 2008, 12:37
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