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.

Answer: E.

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.

You have solved my problem here. i think i can handle these questions now