Use the graph of [tex]\( f(x) = \begin{cases}
x + 3 & \text{if } x \ \textless \ -1 \\
-4x - 2 & \text{if } x \geq -1
\end{cases} \)[/tex]

Over what part of the domain is function [tex]\( f \)[/tex] increasing?

A. [tex]\( x \ \textless \ -1 \)[/tex]

B. [tex]\( x \geq 2 \)[/tex]

C. [tex]\( x \geq -1 \)[/tex]

D. [tex]\( x \ \textless \ 2 \)[/tex]



Answer :

To determine over which part of the domain the function [tex]\( f(x) \)[/tex] is increasing, we need to analyze the behavior of the function's pieces and their respective intervals.

The function [tex]\( f(x) \)[/tex] is defined as:
[tex]\[ f(x) = \left\{ \begin{array}{l} x + 3 \quad \text{for} \; x < -1 \\ -4x - 2 \quad \text{for} \; x \geq -1 \end{array} \right. \][/tex]

### Step-by-Step Analysis:

1. For [tex]\( x < -1 \)[/tex]:

Here, the function is defined as [tex]\( f(x) = x + 3 \)[/tex].

- To determine if this part is increasing, we'll consider the first derivative of [tex]\( f(x) = x + 3 \)[/tex]:
[tex]\[ \frac{d}{dx}(x + 3) = 1 \][/tex]
- Since the derivative is positive (1 > 0) for all [tex]\( x < -1 \)[/tex], the function is indeed increasing in this interval.

2. For [tex]\( x \geq -1 \)[/tex]:

Here, the function is given by [tex]\( f(x) = -4x - 2 \)[/tex].

- To determine if this part is increasing, consider the first derivative of [tex]\( f(x) = -4x - 2 \)[/tex]:
[tex]\[ \frac{d}{dx}(-4x - 2) = -4 \][/tex]
- Since the derivative is negative (-4 < 0) for all [tex]\( x \geq -1 \)[/tex], the function is decreasing in this interval.

### Conclusion:

From the analysis, we observe that the function [tex]\( f(x) \)[/tex] is increasing in the interval where [tex]\( x < -1 \)[/tex]. Therefore, the correct part of the domain over which the function [tex]\( f \)[/tex] is increasing is:

Option A: [tex]\( x < -1 \)[/tex]