Given the function $$ p(x)=\left\{\begin{array}{ll}4 x+7 & x \leq 0 \ -3 x^{2}-3 x+7 & x0\end{array} ight. $$

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Given the function $$ p(x)=\left\{\begin{array}{ll}4 x+7 & x \leq 0 \ -3 x^{2}-3 x+7 & x>0\end{array}ight. $$

UpStudy AI · Accepted Answer

Certainly! Let's analyze the given piecewise function $$ p(x) $$:

$$
p(x) = \begin{cases} 
4x + 7 & \text{if } x \leq 0, \
-3x^{2} - 3x + 7 & \text{if } x > 0.
\end{cases}
$$

### 1. **Understanding the Function**

The function $$ p(x) $$ is defined differently for values of $$ x $$ less than or equal to 0 and for values greater than 0:

- **For $$ x \leq 0 $$:** $$ p(x) $$ is a **linear function** $$ 4x + 7 $$.
- **For $$ x > 0 $$:** $$ p(x) $$ is a **quadratic function** $$ -3x^{2} - 3x + 7 $$.

### 2. **Graphical Representation**

- **Linear Part ($$ x \leq 0 $$):**
  - **Slope:** 4
  - **Y-intercept:** 7
  - **Behavior:** Increasing linearly as $$ x $$ decreases.

- **Quadratic Part ($$ x > 0 $$):**
  - **Coefficient of $$ x^{2} $$:** -3 (indicates a downward-opening parabola)
  - **Vertex:** The parabola opens downward, so it has a maximum point.

### 3. **Continuity at $$ x = 0 $$**

To check if $$ p(x) $$ is continuous at $$ x = 0 $$, evaluate the left-hand limit, right-hand limit, and the function value at $$ x = 0 $$:

- **Left-Hand Limit ($$ x \to 0^- $$):**
  $$
  \lim_{x \to 0^-} p(x) = 4(0) + 7 = 7
  $$
  
- **Right-Hand Limit ($$ x \to 0^+ $$):**
  $$
  \lim_{x \to 0^+} p(x) = -3(0)^2 - 3(0) + 7 = 7
  $$
  
- **Function Value at $$ x = 0 $$:**
  $$
  p(0) = 4(0) + 7 = 7
  $$
  
Since all three values are equal, $$ p(x) $$ is **continuous at $$ x = 0 $$**.

### 4. **Differentiability at $$ x = 0 $$**

To determine if $$ p(x) $$ is differentiable at $$ x = 0 $$, we examine the derivatives from the left and right:

- **Derivative for $$ x \leq 0 $$:** 
  $$
  p'(x) = 4
  $$
  
- **Derivative for $$ x > 0 $$:** 
  $$
  p'(x) = \frac{d}{dx}(-3x^{2} - 3x + 7) = -6x - 3
  $$
  
- **Left-Hand Derivative at $$ x = 0 $$:** 
  $$
  p'_-(0) = 4
  $$
  
- **Right-Hand Derivative at $$ x = 0 $$:** 
  $$
  p'_+(0) = -6(0) - 3 = -3
  $$
  
Since $$ p'_-(0) \neq p'_+(0) $$ (i.e., $$ 4 \neq -3 $$), the function $$ p(x) $$ is **not differentiable at $$ x = 0 $$**.

### 5. **Finding the X-Intercepts**

- **For $$ x \leq 0 $$:** Solve $$ 4x + 7 = 0 $$:
  $$
  x = -\frac{7}{4} = -1.75
  $$
  
- **For $$ x > 0 $$:** Solve $$ -3x^{2} - 3x + 7 = 0 $$:
  $$
  -3x^{2} - 3x + 7 = 0 \
  \Rightarrow 3x^{2} + 3x - 7 = 0 \
  $$
  Using the quadratic formula:
  $$
  x = \frac{-3 \pm \sqrt{9 + 84}}{6} = \frac{-3 \pm \sqrt{93}}{6}
  $$
  $$
  \sqrt{93} \approx 9.6437 \
  x \approx \frac{-3 + 9.6437}{6} \approx 1.107 \quad (\text{valid since } x > 0) \
  x \approx \frac{-3 - 9.6437}{6} \approx -2.4406 \quad (\text{discarded since } x > 0)
  $$
  
  So, the **x-intercepts** are at:
  $$
  x = -1.75 \quad \text{and} \quad x \approx 1.107
  $$

### 6. **Behavior of the Function**

- **For $$ x \leq 0 $$:**
  - The function is a straight line increasing as $$ x $$ increases towards 0.
  
- **For $$ x > 0 $$:**
  - The function is a downward-opening parabola with a maximum value at its vertex.
  - **Vertex of the Parabola:**
    $$
    x = -\frac{b}{2a} = -\frac{-3}{2(-3)} = -\frac{3}{-6} = 0.5
    $$
    $$
    p(0.5) = -3(0.5)^2 - 3(0.5) + 7 = -0.75 - 1.5 + 7 = 4.75
    $$
    
  - The function reaches its maximum at $$ x = 0.5 $$ with $$ p(0.5) = 4.75 $$ and decreases on either side of this vertex.

### 7. **Domain and Range**

- **Domain:** All real numbers $$ (-\infty, \infty) $$.
  
- **Range:**
  - For $$ x \leq 0 $$, as $$ x \to -\infty $$, $$ p(x) \to -\infty $$.
  - For $$ x > 0 $$, as $$ x \to \infty $$, $$ p(x) \to -\infty $$.
  - The maximum value of $$ p(x) $$ is 7 (achieved at $$ x = 0 $$).

Therefore, the **range** is $$ (-\infty, 7] $$.

### 8. **Summary**

- **Continuity:** The function $$ p(x) $$ is continuous at $$ x = 0 $$.
- **Differentiability:** $$ p(x) $$ is **not** differentiable at $$ x = 0 $$ due to a sharp corner (discontinuity in the derivative).
- **Graph:** Consists of a straight line for $$ x \leq 0 $$ and a downward-opening parabola for $$ x > 0 $$, smoothly connected at $$ x = 0 $$.
- **Intercepts:** 
  - Y-intercept at $$ (0, 7) $$.
  - X-intercepts at $$ (-1.75, 0) $$ and approximately $$ (1.107, 0) $$.

If you have a specific aspect of $$ p(x) $$ you'd like to explore further—such as solving for specific values, integrating, differentiating elsewhere, or graphing—feel free to ask!
The function $$ p(x) $$ is defined as: $$ p(x) = \begin{cases} 4x + 7 & \text{if } x \leq 0, \ -3x^{2} - 3x + 7 & \text{if } x > 0. \end{cases} $$ - **Continuity:** $$ p(x) $$ is continuous at $$ x = 0 $$. - **Differentiability:** Not differentiable at $$ x = 0 $$ due to a sharp corner. - **Intercepts:** Y-intercept at $$ (0, 7) $$, x-intercepts at $$ (-1.75, 0) $$ and approximately $$ (1.107, 0) $$. - **Range:** $$ (-\infty, 7] $$. The function consists of a straight line for $$ x \leq 0 $$ and a downward-opening parabola for $$ x > 0 $$, smoothly connected at $$ x = 0 $$.

Given the function \( p(x)=\left\{\begin{array}{ll}4 x+7 & x \leq 0 \\ -3 x^{2}-3 x+7 & x>0\end{array}\right. \)

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