Fie functia f : R −→ R, definita prin f (x) = x + 1, ∀x ∈ R. Sa se rezolve ecuatia(f ◦ f )(x) = f 2(x).

1. Break Down the Problem

We need to solve the equation $(f \circ f)(x) = f^2(x)$.

1. Determine $f \circ f$, which means applying the function $f$ twice.
2. Determine $f^2(x)$, which we will interpret as applying the function $f$ two times as well.
3. Set the results of these operations equal and solve for $x$.

2. Relevant Concepts

The function is defined as: $f(x) = x + 1$

1. $f \circ f(x)$ means: $f(f(x)) = f(x + 1)$
2. $f^2(x)$ means: $f(f(x)) = f(x + 1) = (x + 1) + 1$

3. Analysis and Detail

Let's compute $f(f(x))$ and $f^2(x)$:

1. Compute $f \circ f(x)$: $f(f(x)) = f(x + 1) = (x + 1) + 1 = x + 2$

2. Compute $f^2(x)$: $f^2(x) = f(f(x)) = (f(x + 1)) = (x + 1) + 1 = x + 2$

Now we have: $(f \circ f)(x) = x + 2$ $f^2(x) = x + 2$

Thus the equation becomes: $x + 2 = x + 2$

4. Verify and Summarize

The equation $x + 2 = x + 2$ is valid for all $x \in R$. Therefore, every real number $x$ satisfies the equation.

Final Answer

The solution to the equation is: $\text{Solution:} \quad x \in \mathbb{R}$