SOLUTION: f(a+b) = f(a)+ f(b) for all positive numbers a and b
a) f(x) = x squared
b) f(x) = x+1
c) f(x) = square root x
d) f(x) = 2/x
e) f(x) = -3x
Thanks,
Casey
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-> SOLUTION: f(a+b) = f(a)+ f(b) for all positive numbers a and b
a) f(x) = x squared
b) f(x) = x+1
c) f(x) = square root x
d) f(x) = 2/x
e) f(x) = -3x
Thanks,
Casey
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Question 74102: f(a+b) = f(a)+ f(b) for all positive numbers a and b
a) f(x) = x squared
b) f(x) = x+1
c) f(x) = square root x
d) f(x) = 2/x
e) f(x) = -3x
Thanks,
Casey Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! 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) 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) Which is not equal, that eliminates b)
c) Which is not equal, so that eliminates c)
d) Not equal, that eliminates d). There's only one left, lets verify if this is our answer.
e) Which works. Notice how the two sides are simply the right side has distributed the -3 where the left side factored the -3