Let's calculate each value by substituting the respective values of $x$ into the function $f(x) = \frac{2x + 4}{x + 1}$.

1. For $f(-2)$: $f(-2) = \frac{2(-2) + 4}{-2 + 1} = \frac{-4 + 4}{-1} = \frac{0}{-1} = 0$ So, $f(-2) = 0$.

2. For $f(-1)$: $f(-1) = \frac{2(-1) + 4}{-1 + 1} = \frac{-2 + 4}{0} = \frac{2}{0}$ Since division by zero is undefined, $f(-1) = \text{DNE}$.

3. For $f(0)$: $f(0) = \frac{2(0) + 4}{0 + 1} = \frac{0 + 4}{1} = 4$ So, $f(0) = 4$.

4. For $f(1)$: $f(1) = \frac{2(1) + 4}{1 + 1} = \frac{2 + 4}{2} = \frac{6}{2} = 3$ So, $f(1) = 3$.

5. For $f(2)$: $f(2) = \frac{2(2) + 4}{2 + 1} = \frac{4 + 4}{3} = \frac{8}{3} \approx 2.67$ So, $f(2) \approx 2.67$ (rounded to two decimal places).

Summary of Results:

• $f(-2) = 0$
• $f(-1) = \text{DNE}$
• $f(0) = 4$
• $f(1) = 3$
• $f(2) \approx 2.67$

Would you like a detailed explanation of any specific step, or have further questions?

1. What is the process to evaluate functions at specific points?
2. How do undefined values affect the function's domain?
3. Can a function have multiple undefined points?
4. What types of functions often lead to undefined points?
5. How do you approximate irrational results to two decimal places?

Tip: Always check the denominator in rational functions to avoid undefined values.