Question

Question
For the piecewise function below, which of the following statements is true?

math expressionf left parenthesis x right parenthesis equals enlarged left brace start layout 1st row left parenthesis x plus 3 right parenthesis squared minus 1 comma negative 5 less than or equals x less than or equals negative 1 2nd row negative x comma negative 1 less than x less than or equals 1 3rd row start fraction 5 over 2 end fraction log subscript 2 baseline left parenthesis negative x plus 4 right parenthesis minus 1 comma 1 less than x less than or equals 4 end layout
(1 point)
Responses

f left parenthesis negative 1 right parenthesis greater than f left parenthesis 1 right parenthesis
Image with alt text: f left parenthesis negative 1 right parenthesis greater than f left parenthesis 1 right parenthesis

f left parenthesis negative 1 right parenthesis less than f left parenthesis 1 right parenthesis
Image with alt text: f left parenthesis negative 1 right parenthesis less than f left parenthesis 1 right parenthesis

f left parenthesis negative 1 right parenthesis equals f left parenthesis 1 right parenthesis
Image with alt text: f left parenthesis negative 1 right parenthesis equals f left parenthesis 1 right parenthesis

The function is not defined at both f left parenthesis negative 1 right parenthesis and f left parenthesis 1 right parenthesis.
The function is not defined at both Image with alt text: f left parenthesis negative 1 right parenthesis and Image with alt text: f left parenthesis 1 right parenthesis .
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Answer 1

To determine which statement about the piecewise function \( f(x) \) is true, we need to evaluate \( f(-1) \) and \( f(1) \) based on the defined intervals of the function.

The piecewise function is defined as follows:

1. \( f(x) = (x + 3)^2 - 1 \) for \( -5 \leq x \leq -1 \)
2. \( f(x) = -x \) for \( -1 < x \leq 1 \)
3. \( f(x) = \frac{5}{2} \log_2(-x + 4) - 1 \) for \( 1 < x \leq 4 \)

Now, let's evaluate \( f(-1) \) and \( f(1) \):

### Evaluate \( f(-1) \):
Since \( -1 \) falls on the boundary of the first interval, we will use the first piece:
\[
f(-1) = (-1 + 3)^2 - 1 = (2)^2 - 1 = 4 - 1 = 3
\]

### Evaluate \( f(1) \):
Since \( 1 \) also falls on the boundary of the second interval, we use the second piece:
\[
f(1) = -1
\]

### Compare \( f(-1) \) and \( f(1) \):
Now we compare the two values we calculated:
\[
f(-1) = 3 \quad \text{and} \quad f(1) = -1
\]
Thus, \( f(-1) > f(1) \).

### Conclusion:
The correct statement is:
\[
f(-1) > f(1)
\]

So the answer is:
- \( f(-1) > f(1) \)