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The given problem is to show that the function \( h(x) \), defined piecewise, is affine (i.e., linear with a constant) by intervals. Let's break down the function:

\[
h(x) =
\begin{cases}
1 + \frac{x^2 - xE(x)}{x + 1} & \text{if } x \leq 0 \\
f(x) & \text{if } x > 0
\end{cases}
\]

Where \( E(x) \) likely refers to the floor function, which gives the greatest integer less than or equal to \( x \).

### Steps to solve the problem:

1. **For \( x \leq 0 \):** We need to simplify the first part of the function \( 1 + \frac{x^2 - xE(x)}{x + 1} \). Since \( x \leq 0 \), \( E(x) = x \), because the floor of \( x \) when \( x \) is an integer or less is \( x \) itself. Therefore, we can rewrite the expression as:

   \[
   1 + \frac{x^2 - x \cdot x}{x + 1} = 1 + \frac{x^2 - x^2}{x + 1} = 1
   \]

   Hence, for \( x \leq 0 \), \( h(x) = 1 \), which is a constant function (affine with a slope of 0).

2. **For \( x > 0 \):** The function is defined as \( f(x) \), but since we don't have the explicit form of \( f(x) \), we cannot simplify it directly. However, we need to check if \( f(x) \) is affine in this interval (for some definition of \( f(x) \)). This might require additional information about \( f(x) \).

### Conclusion:
We have shown that for \( x \leq 0 \), the function is constant, which is a form of an affine function. To complete the solution for \( x > 0 \), we would need the form of \( f(x) \).

Would you like further clarification or details on any step? 

Here are some questions that could expand your understanding:
1. Can we assume a specific form of \( f(x) \) for further analysis?
2. What is the behavior of \( h(x) \) around \( x = 0 \)?
3. How does the floor function \( E(x) \) affect the result for non-integer \( x \)?
4. What are the general conditions for a function to be affine?
5. How would we approach showing that \( f(x) \) is affine?

**Tip**: For piecewise functions, it's essential to analyze each interval separately and check for continuity or other properties at the boundaries.

Soit h une fonction définie sur ]-1,+∞[ par : h(x) = {1 + (x^2 - xE(x))/(x + 1) si x ≤ 0; f(x) si x > 0}. Montrer que h est une fonction affine par intervalles.

Analyze the piecewise function h(x) defined on ]-1,+∞[ to show that it is affine by intervals. Explore the impact of the floor function and learn to identify affine properties in each interval.

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