Question

If f(x1)+f(x2)=f(x1+x2) for all real numbers x1 and x2, which of the following could define f?

A) f(x)=x+1
B)f(x)=2x
C)f(x)=1/x
D)f(x)=e^x
E)f(x)=x^2

I have no idea how to go about this, if someone can explain it to me.

Answer 1

Answer with explanation:

For a function f(x), it is given that following Statement hold.

`f(x_(1))+f(x_(2))=f(x_(1)+x_(2))`

Coming to the options

(A)f(x)=x+1

`f(x_(1))=x_(1)+1\\\\f(x_(2))=x_(2)+1\\\\f(x_(1)+x_(2))=x_(1)+x_(2)+1\\eq f(x_(1))+f(x_(2))`

This function does not define f.

(B)f(x)=2x

`f(x_(1))=2x_(1)\\\\f(x_(2))=2x_(2)\\\\f(x_(1)+x_(2))=2(x_(1)+x_(2))\\\\=2x_(1)+2x_(2)=f(x_(1))+f(x_(2))`

This function defines f.

`(C)\rightarrow f(x)=(1)/(x)\\\\f(x_(1))=(1)/(x_(1))\\\\f(x_(2))=(1)/(x_(2))\\\\f(x_(1))+f(x_(2))=(1)/(x_(2))+(1)/(x_(1))\\\\f(x_(1)+x_(2))=(1)/(x_(1)+x_(2))\\\\f(x_(1))+f(x_(2))\\eq f(x_(1)+x_(2))`

This function does not define f.

`(D)\rightarrow f(x)=e^x\\\\f(x_(1))=e^{x_(1)}\\\\f(x_(2))=e^{x_(2)}\\\\f(x_(1))+f(x_(2))=e^{x_(2)}+e^{x_(1)}\\\\f(x_(1)+x_(2))=e^{x_(1)+x_(2)}\\eq f(x_(1))+f(x_(2))`

This function does not define f.

`(E)\rightarrow f(x)=x^2\\\\f(x_(1))=(x_(1))^2\\\\f(x_(2))=(x_(2))^2\\\\f(x_(1))+f(x_(2))=(x_(1))^2+(x_(2))^2\\\\f(x_(1)+x_(2))=[x_(1)+x_(2)]^2\\eq f(x_(1))+f(x_(2))`

This function does not define f.

Option B: f(x)=2x , defines f.

Answer 2

For the answer to the question above, the answer is simple.
it is multiple choice letter B since f(x1) + f(x2) = 2x1 + 2x2 = 2(x1+x2) = f(x1+x2)
I hope my answer helped you in your problem, Have a nice day!