find f(x) = x + 1 if x < 2
find f(x) = x² if x ≥ -1

The piecewise function \( f(x) \) you've described is defined differently depending on the value of \( x \). However, there seems to be a contradiction in the intervals provided:

1. \( f(x) = x + 1 \) if \( x < 2 \)
2. \( f(x) = x^2 \) if \( x \geq -1 \)

The issue is that the second condition \( x \geq -1 \) overlaps with the first condition \( x < 2 \). Therefore, we need to clarify how to define \( f(x) \) for \( -1 \leq x < 2 \) where both conditions could apply.

One possible interpretation is to consider that the first condition takes priority for values of \( x \) less than 2, and the second condition takes priority for values of \( x \) greater than or equal to 2. 

So, the piecewise function can be defined as:
\[
f(x) =
\begin{cases} 
x + 1 & \text{if } x < 2 \\
x^2 & \text{if } x \geq 2 
\end{cases}
\]

This interpretation assumes that \( f(x) = x^2 \) only applies strictly when \( x \geq 2 \). If instead, you meant \( f(x) = x^2 \) to apply as soon as \( x \geq -1 \), you would need to define how the function behaves within the overlapping interval \( -1 \leq x < 2 \).

Would you like to discuss this further or need clarification?

### Related Questions:
1. What is the value of \( f(x) \) at \( x = 0 \)?
2. How would you graph this piecewise function?
3. What is the value of \( f(x) \) at \( x = 2 \)?
4. Can you find the derivative of this piecewise function?
5. How does this piecewise function behave as \( x \) approaches \( 2 \)?

**Tip:** When dealing with piecewise functions, ensure the intervals do not overlap, or clearly define which condition applies within overlapping intervals.

Explore piecewise functions with an emphasis on definition and examples. This problem involves defining a piecewise function where conditions overlap, requiring careful interpretation. Suitable for Grades 10-12.

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