Question 74102
If we let a=2 and b=3 (arbitrary numbers) and we plug them into each possible equation, we can eliminate the wrong answers.

a)Lets start with our first possible answer.
f(a+b) = f(a)+ f(b) Plug in a=2,b=3
f(2+3)=f(2)+f(3)
{{{(2+3)^2=2^2+3^2}}}
{{{25=4+9}}}
{{{25=9}}}Which is obviously not equal, so that eliminates a) continue with the rest by plugging in a+b in for x on the left and a and b separately for x on the right.


b)
{{{(2+3)+1=(2+1)+(3+1)}}}
{{{6=7}}}Which is not equal, that eliminates b)


c)
{{{sqrt(2+3)=sqrt(2)+sqrt(3)}}}
{{{sqrt(5)=1.414+1.732}}}
{{{2.236=3.146}}}Which is not equal, so that eliminates c)


d)
{{{2/(2+3)=2/2+2/3}}}
{{{2/5=1+2/3}}}
{{{2/5=5/3}}}Not equal, that eliminates d). There's only one left, lets verify if this is our answer.


e)
{{{-3(2+3)=-3(2)+(-3(3))}}}
{{{-3(5)=-6-9}}}
{{{-15=-15}}}Which works. Notice how the two sides are simply the right side has distributed the -3 where the left side factored the -3