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For which of the following functions is f(a+b)=f(a)+f(b) for

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23 Apr 2010, 15:24
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For which of the following functions is f(a+b)=f(a)+f(b) for all positive numbers a and b?

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$
[Reveal] Spoiler: OA
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24 Apr 2010, 06:02
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hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

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

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Similar questions:
for-which-of-the-following-functions-is-f-a-b-f-b-f-a-124491.html
for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
let-the-function-g-a-b-f-a-f-b-143311.html

Hope it helps.
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24 Apr 2010, 09:23
Thank you Bunuel. This is really a good explaination. I did some samples based on your explaination and i am confident that I can handle these kind or problems.
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24 Apr 2010, 09:36
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Bunuel wrote:
hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

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

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

Thanks bunuel
You have solved my problem here. i think i can handle these questions now
+1
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Re: functions Problem need help. [#permalink]

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24 Apr 2010, 14:51
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Bunuel wrote:
hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

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

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

thanks a ton for the explanation. There are not many problems on functions in samples and I am glad now i know how to go about for questions ike this.
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15 Jun 2010, 05:12
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For these kind of questions where you have to test each choice, ALWAYS start with E.

E. f(x) = -3x
f(a+b) = -3(a+b) = -3a -3b
f(a) = -3a ; f(b) = -3b
f(a) + f(b)= -3a -3b = f(a+b)

On test day - stop.

Remember, just substitute for x with whatever is in the brackets in f ( )
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15 Jun 2010, 05:22
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I agree with AbhayPrasanna.

If you want an explanation of why it doesn't work for the other functions, look at the type of each function.

$$x^2$$ is a quadratic function, so $$f(a+b) = (a+b)^2$$ and $$f(a) + f(b) = a^2 + b^2$$ - Wrong

$$x+1$$ has a constant variable in it, though its linear. So $$f(a+b) = a+b+1$$ and $$f(a) + f(b) = (a+1)+(b+1) = a+b+2$$ - Wrong

$$\sqrt{x}$$ is a root function. $$f(a+b) = \sqrt{a+b}$$ and $$f(a)+f(b) = \sqrt{a}+\sqrt{b}$$ - Wrong

$$2/x$$ is a fraction type function. So $$f(a+b) = 2/(a+b)$$ and $$f(a)+f(b) = 2/a + 2/b = 2(a+b)/ab$$ - Wrong

$$-3x$$ is a linear function without constants. So this must be the answer. But to check:

$$f(a+b) = -3*(a+b)$$ and $$f(a) + f(b) = -3*a + (-3*b) = -3* (a+b)$$ - Right

This kind of elimination is not necessary on the exam. However, it's good to know why the other options don't work.
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30 Dec 2010, 01:38
i think Karishma has explained very good point to tackle these functions problems, As per her explanation we should first try options with multiple , divide add and then subtract…. So trying option E was obvious and it fits well… saves time indeed…

Ans E
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30 Dec 2010, 14:41
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marijose wrote:
whiplash2411 wrote:
I agree with AbhayPrasanna.

If you want an explanation of why it doesn't work for the other functions, look at the type of each function.

$$x^2$$ is a quadratic function, so $$f(a+b) = (a+b)^2$$ and $$f(a) + f(b) = a^2 + b^2$$ - Wrong

$$x+1$$ has a constant variable in it, though its linear. So $$f(a+b) = a+b+1$$ and $$f(a) + f(b) = (a+1)+(b+1) = a+b+2$$ - Wrong

$$\sqrt{x}$$ is a root function. $$f(a+b) = \sqrt{a+b}$$ and $$f(a)+f(b) = \sqrt{a}+\sqrt{b}$$ - Wrong

$$2/x$$ is a fraction type function. So $$f(a+b) = 2/(a+b)$$ and $$f(a)+f(b) = 2/a + 2/b = 2(a+b)/ab$$ - Wrong

$$-3x$$ is a linear function without constants. So this must be the answer. But to check:

$$f(a+b) = -3*(a+b)$$ and $$f(a) + f(b) = -3*a + (-3*b) = -3* (a+b)$$ - Right

This kind of elimination is not necessary on the exam. However, it's good to know why the other options don't work.

Is there a way to solve the problem without doing all the calculations?

Basically there are 2 approaches possible: algebraic and number plugging, check my post for both.
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03 Jan 2011, 01:38
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Merging similar topics.

Another question on the same concept with solution at: function-85751.html?hilit=following%20functions%20positive#p644387

Hope it helps
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03 Jan 2011, 19:52
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Bunuel wrote:
hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

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

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

Thanks for really fleshing out the algebra on this problem Bunuel. The problems seem fairly easy once you understand how to work functions properly.
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21 Jun 2013, 02:46
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Re: For which of the following functions is f(a+b)=f(a)+f(b) for [#permalink]

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22 Jun 2013, 00:44
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Well, i did not start with picking no's or keeping the no's as a and b,

But just worked in the following way :-
if, f(a+b) = f(a) + f(b)
Then,
since our answers are in this format,

f(x+x) = f(x) + f(x) should also hold true, for all positive no's a and b (we havent been told that it has to be distinct, so this should be perfectly valid)

Hence, you need to evaluate f(2x) here, and check weather it is equal to 2f(x)

This seems to be the case only for E, hence, the answer.
Quick and easy.
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Re: For which of the following functions is f(a+b)=f(a)+f(b) for [#permalink]

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30 Sep 2014, 00:21
zz0vlb wrote:
For which of the following functions is f(a+b)=f(a)+f(b) for all positive numbers a and b?

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

f(a+b) = f(a) + f(b) holds true for pure multiplication with any number

D: $$\frac{2}{x} = 2 * x^{-1}$$ ......... power function.... discard

E: Only Multiplication.... This will be the answer

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