For which of the following functions is $f(a+b) = f(a) + f(b)$?

Question

For which of the following functions $f(x)$ is $f(a+b) = f(a) + f(b)$?

(A) $f(x) = x^2$
(B) $f(x) = 5x$
(C) $f(x) = 2x+1$
(D) $f(x) = \sqrt{x}$
(E) $f(x) = x-2$

Given solution:

Let $a = 2$, and $b = 3$.

Then solve each of the given functions like the following:

the snap shot suggests that I should calculate all functions which I don't like.

Is there any better method out there?

Answer 1

( not better )

$$ f\left(a+b\right)=f\left(a\right)+f\left(b\right) $$ For $a=b$ we have $$ f\left(2a\right)=2f\left(a\right) $$ Taking $a=0$ gives you $f(0)=0$ and for $b=-a$ we have $f(0)=0=f(x)+f(-x)$ so $f$ is odd. Then you can show by induction that $$ f(n)=nf(1) $$ Then $f$ is odd so it is true for $n \in \mathbb{Z}$. With $r=p/q \in \mathbb{Q}$ we have $$ f\left(r\right)=f\left(\frac{p}{q}\right)=pf\left(\frac{1}{q}\right) \text{ and }f\left(1\right)=f\left(\frac{q}{q}\right)=qf\left(\frac{1}{q}\right) $$ So $$ f\left(r\right)=\frac{p}{q}f\left(1\right)=rf\left(1\right) $$ Then $Q$ is dense in $\mathbb{R}$ so $f$ satisfies $$ f\left(x\right)=xf\left(1\right) $$ So the only solution is $f(x)=5x$ in your exercise.

Answer 2

To elaborate on Gabriel Romon's comment, a function $f$ that satisfies $$ f(x + y) = f(x) + f(y) $$ is called linear. This definition immediately restricts any function different from $f(x) = ax + b$, as you know from basic calculus that if $x$ has any power other then $1$ or $0$, then it's not linear (in the normal sense). This eliminates $(a)$ and $(d)$ immediately.

It remains to see if $f(x) = ax + b$ is actually linear, in this new sense. Quickly check that $$ f(x + y) = a(x + y) + b = ax + b + y = f(x) + y \neq f(x) + f(y), $$ so the constant $b$ stops $f$ from being linear. This means that all you're left with is $f(x) = ax$, so only option $(b)$ satisfies this.

This is a very 'hand wavey' argument, but I think it suffices for your purposes.

Answer 3

All constants will be counted only once in $f(a+b)$ but twice in $f(a)+f(b)$, so any functions with a constant are automatically excluded. In this case, thats is $C$ and $E$.

When raising to powers, note that $(a+b)^n\ne a^n+b^n$ unless $n=1$, $a=0$, or $b=0$, thus any other powers won't work. Thus $A$ and $D$ are ruled out. When $n=1$, then $k(a+b)^n=k(a^n+b^n)$, and so a function of the form $y=kx$ is perfectly valid, hence $B$ is correct.

Answer 4

I wouldn't solve any of them.

• I searched around a bit and reminded myself that the key here is the "Distributive Property (of Multiplication) over Addition".

  https://www.khanacademy.org/math/pre-algebra/pre-algebra-arith-prop/pre-algebra-ditributive-property/v/the-distributive-property

  This is why B will work.

• There is a distributive property over exponentiation, but it distributes over multiplication, not over addition.

  http://www.solving-math-problems.com/exponent-rules-distributive.html

  This is why A and D will not work.

• There is no distributive property (of addition) over addition.

  This is why C and E will not work.

You asked if there was another way. Knowing why things work instead of just how to do the problems is a very good way.

Answer 5

In such type of problems go by option verification.

A)$f(a+b)=(a+b)^2=a^2+b^2+2ab\neq f(a)+f(b)$

B)$f(a+b)=5(a+b)=5a+5b=f(a)+f(b)$

and so on$.....$ for other options

So, the option is $B$

OR You can save time by using an intuitive method. Look for the expression that satisfies the distributive property $i.e.$ $x (y + z) = xy + x z$

When you put $(a+b)$, it should give individual functions in $a$ and $b$ which means that you will get two separate, comparable terms in $a$ and $b$. Squares, roots, addition and division by the variable does not satisfy the distributive property.