Find $f(3)$ if $f(f(x))=3+2x$

Question

A function $f\colon \mathbb{R} \to \mathbb{R}$ is defined as $f(f(x))=3+2x$

Find $f(3)$ if $f(0)=3$

My try:

Method $1.$ Put $x=0$ we get $f(f(0))=3$ $\implies$ $f(3)=3$

Method $2.$ Replace $x$ with $f(x)$ we get

$$f(f(f(x)))=3+2f(x)$$ $\implies$

$$f(3+2x)=3+2f(x)$$

Put $x=0$

$$f(3)=9$$

I feel Method $2.$ is Correct since $f(f(x))=3+2x$ is Injective which means $f(x)$ should be Injective.

Answer 1

It looks like you have found a contradiction. Apparently, $f(f(x)) = 3 + 2x$ and $f(0) = 3$ are not reconcilable. There is no such function.

Answer 2

Not only is $f(f(x))=3+2x$ injective, it is bijective, and $f(f(x))\not=x$ except when $x=-3$

so you must have $f(x)\not =x$ and thus must have have $f(f(x)) \not =f(x)$, except for the case $x=-3$ in which case you must have $f(-3)=-3$

in particular you cannot have $f(0)=f(f(0))=3$

In fact you can have $f(0)$ with any values apart from $-3$ or those in $\left\{\ldots, -\frac{45}{16},-\frac{21}{8},-\frac{9}{4},-\frac{3}{2},0,3,9,21,45,\ldots\right\}$, and you will then have $f(3)=3+2f(0)$

For a general solution to the functional equation,

• you partition the reals into equivalence classes with the equivalence relation $y = 3+2x \implies y R x$
• then pair these equivalence classes up, apart from $\{-3\}$ as a special case paired with itself, (there are some easy ways of doing this, and many more difficult ones)
• so $f(x)=z$ takes you from one equivalence class to its pair
• and $f(z)=f(f(x))=3+2x$ takes you back but one step up

One example solution to the functional equation is $f(x)=-6-x$ when $x\ge -3$ and $f(x)=-9-2x$ when $x\le -3$, which has $f(0)=-6$ and $f(3)=-9$