$$ q(x)=\left\{\begin{array}{ll}-3 x-9 & ext { if } x \leq-3 \ \frac{5}{9} x^{2} & ext { if } x-3\end{array} ight. $$

Question

$$ q(x)=\left\{\begin{array}{ll}-3 x-9 & 	ext { if } x \leq-3 \ \frac{5}{9} x^{2} & 	ext { if } x>-3\end{array}ight. $$

UpStudy AI · Accepted Answer

The function $$ q(x) $$ is defined piecewise, meaning it has different expressions based on the value of $$ x $$. Let's break it down:

1. **For $$ x \leq -3 $$**:
   $$
   q(x) = -3x - 9
   $$

2. **For $$ x > -3 $$**:
   $$
   q(x) = \frac{5}{9}x^2
   $$

### Analyzing the Function

To analyze the function, we can look at several aspects:

- **Domain**: The function is defined for all real numbers since it has conditions for both $$ x \leq -3 $$ and $$ x > -3 $$.

- **Continuity**: We need to check if the function is continuous at the point where the definition changes, which is at $$ x = -3 $$.

### Step 1: Check Continuity at $$ x = -3 $$

To check continuity at $$ x = -3 $$, we need to find:
- $$ q(-3) $$ using the first piece of the function.
- The limit of $$ q(x) $$ as $$ x $$ approaches -3 from the left and right.

#### Calculate $$ q(-3) $$:
Using the first piece:
$$
q(-3) = -3(-3) - 9 = 9 - 9 = 0
$$

#### Calculate the left-hand limit as $$ x $$ approaches -3:
Using the first piece:
$$
\lim_{x \to -3^-} q(x) = -3(-3) - 9 = 0
$$

#### Calculate the right-hand limit as $$ x $$ approaches -3:
Using the second piece:
$$
\lim_{x \to -3^+} q(x) = \frac{5}{9}(-3)^2 = \frac{5}{9} \cdot 9 = 5
$$

### Step 2: Conclusion on Continuity

Since:
- $$ q(-3) = 0 $$
- $$ \lim_{x \to -3^-} q(x) = 0 $$
- $$ \lim_{x \to -3^+} q(x) = 5 $$

The left-hand limit does not equal the right-hand limit, hence $$ q(x) $$ is not continuous at $$ x = -3 $$.

### Summary

- The function $$ q(x) $$ is defined piecewise with two different expressions.
- It is not continuous at $$ x = -3 $$ because the left-hand limit and right-hand limit do not match.

If you have any specific questions or need further analysis, feel free to ask!
The function $$ q(x) $$ is defined differently for $$ x \leq -3 $$ and $$ x > -3 $$. It is not continuous at $$ x = -3 $$ because the left-hand and right-hand limits at that point are not equal.

\( q(x)=\left\{\begin{array}{ll}-3 x-9 & \text { if } x \leq-3 \\ \frac{5}{9} x^{2} & \text { if } x>-3\end{array}\right. \)

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