https://file.aiquickdraw.com/mathbot/file/17295252369696lqxz9r2.jpg

Let's solve the given piece-wise function for \( f(0) \).

The function is defined as:

\[
f(x) = 
\begin{cases} 
x + 2 & \text{if } x \leq -1 \\
2x + 4 & \text{if } x > -1 
\end{cases}
\]

We are asked to find \( f(0) \). Since \( 0 > -1 \), we will use the second part of the piece-wise function: 

\[
f(x) = 2x + 4 \quad \text{for} \quad x > -1.
\]

Substitute \( x = 0 \):

\[
f(0) = 2(0) + 4 = 0 + 4 = 4.
\]

So, \( f(0) = 4 \).

Would you like more details on this solution, or do you have any other questions?

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Here are 5 related questions to expand your understanding:
1. How would you evaluate \( f(-2) \) using the same piece-wise function?
2. What happens to the function \( f(x) \) as \( x \) approaches -1 from the left and right?
3. How can you graph this piece-wise function?
4. What would the value of \( f(1) \) be?
5. How would you rewrite this piece-wise function for all real numbers?

**Tip:** When dealing with piece-wise functions, always check which condition your input falls under before substituting into the equation.

Find f(0) for the piece-wise function: f(x) = {x + 2 if x ≤ -1, 2x + 4 if x > -1}

Learn how to evaluate the piecewise function f(x) = {x + 2 if x ≤ -1, 2x + 4 if x > -1} at x = 0 using substitution. This problem is suitable for students in Grades 9-12 and covers key algebraic concepts.

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