Show that a function is $f$ bijective if $f(f(f(2x+3)))=x$ for all real $x$

Question

Let $ f : \mathbb{R} \to \mathbb{R} $ is a function such that
$$ \forall x\in\mathbb{R} : f(f(f(2x+3)))=x $$
Show that $f$ is bijective.

We have to show that $f$ is injective and surjective.
How do we do that when we don't know what $f(x)$ is?
We only have a strange looking equality that holds for all real $x$ and the domain of the function.

Answer 1

Let $y=2x+3$
Then $x=(y-3)/2$

$$ \forall x\in\mathbb{R} : f(f(f(y)))=(y-3)/2 $$

(A) When $y$ covers all real numbers , $(y-3)/2$ will also cover all real numbers.

(B) Let there be 2 $y$ values which have same Image.
That is , $f(y_1)=f(y_2)$
Then we will have $f(f(y_1))=f(f(y_2))$
Eventually , we will have $f(f(f(y_1)))=f(f(f(y_2)))$
That will give $(y_1-3)/2=(y_2-3)/2$
That implies $y_1=y_2$

Answer 2

To add to Prem's answer:

Whenever you have $f(f(...f(x)...)) = g(x)$ where $g$ is bijective, then $f$ must also be bijective, no matter how many times you involute $f$.

Surjective: Since $g$ is surjective (bijective), it takes on all values in $\mathbf{R}$, but we see that everything on the left side is "wrapped" in $f$, and hence $f$ also takes on every value in $\mathbf{R} \implies f$ is surjective.

Injective: Assume $f(a) = f(b)$. Then, $f(...f(a)...) = g(a) = f(...f(b)...) = g(b) \implies g(a) = g(b)$. But since $g$ is injective (bijective), $g(a) = g(b) \iff a = b$.