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

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!

Answer 2

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

 $f(x_{1})=x_{1}+1\\\\f(x_{2})=x_{2}+1\\\\f(x_{1}+x_{2})=x_{1}+x_{2}+1\neq f(x_{1})+f(x_{2})$

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.

$(C)\rightarrow f(x)=\frac{1}{x}\\\\f(x_{1})=\frac{1}{x_{1}}\\\\f(x_{2})=\frac{1}{x_{2}}\\\\f(x_{1})+f(x_{2})=\frac{1}{x_{2}}+\frac{1}{x_{1}}\\\\f(x_{1}+x_{2})=\frac{1}{x_{1}+x_{2}}\\\\f(x_{1})+f(x_{2})\neq f(x_{1}+x_{2})$

This function does not define f.

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

This function does not define f.

$(E)\rightarrow f(x)=x^2\\\\f(x_{1})=(x_{1})^2\\\\f(x_{2})=(x_{2})^2\\\\f(x_{1})+f(x_{2})=(x_{1})^2+(x_{2})^2\\\\f(x_{1}+x_{2})=[x_{1}+x_{2}]^2\neq f(x_{1})+f(x_{2})$

This function does not define f.

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