Question

For the following piecewise functions, evaluate f(_3), f(_2), f(4) and f(2)
F(x)=2x+1, x>0
=3x_1, x<0

Answer 1

It looks like there may be a minor confusion in the definition of the piecewise function \( F(x) \). I'll interpret the function as follows:

\[
F(x) =
\begin{cases}
2x + 1 & \text{if } x > 0 \\
3x & \text{if } x < 0
\end{cases}
\]

Now, we will evaluate \( F(-3) \), \( F(-2) \), \( F(4) \), and \( F(2) \).

1. **Evaluate \( F(-3) \)**:
- Since \(-3 < 0\), we use the second piece.
\[
F(-3) = 3(-3) = -9
\]

2. **Evaluate \( F(-2) \)**:
- Since \(-2 < 0\), we again use the second piece.
\[
F(-2) = 3(-2) = -6
\]

3. **Evaluate \( F(4) \)**:
- Since \(4 > 0\), we use the first piece.
\[
F(4) = 2(4) + 1 = 8 + 1 = 9
\]

4. **Evaluate \( F(2) \)**:
- Since \(2 > 0\), we use the first piece.
\[
F(2) = 2(2) + 1 = 4 + 1 = 5
\]

Now, summarizing the evaluations:
- \( F(-3) = -9 \)
- \( F(-2) = -6 \)
- \( F(4) = 9 \)
- \( F(2) = 5 \)