# Step-by-Step Solution Let's solve the problem step by step. --- ## 1. **Understanding the Piecewise Function** The function \( f(x) \) is defined as: \[ f(x) = \begin{cases} -x^2 + 10 & \text{if } -4 \leq x < 4 \\ 5 - x & \text{if } x \geq 4 \end{cases} \] --- ## 2. **Graph Each Piece** ### **First Piece: \( -x^2 + 10 \) for \( -4 \leq x < 4 \)** - This is a downward-opening parabola. - Domain: \( -4 \leq x < 4 \) - Endpoints: - At \( x = -4 \): \( f(-4) = -(-4)^2 + 10 = -16 + 10 = -6 \) - At \( x = 4 \): \( f(4) = -(4)^2 + 10 = -16 + 10 = -6 \) BUT, since \( x = 4 \) is **not included** (only up to but not including 4), we use an open circle at \( (4, -6) \). - Vertex: \( x=0 \), \( f(0) = -(0)^2 + 10 = 10 \) - Plot points between \( x = -4 \) and \( x = 4 \). ### **Second Piece: \( 5 - x \) for \( x \geq 4 \)** - This is a straight line with slope -1. - Start at \( x = 4 \): \( f(4) = 5 - 4 = 1 \) (included, so solid dot at \( (4, 1) \)) - Continue for \( x > 4 \). --- ## 3. **Check for Continuity at \( x = 4 \)** Let's compare the left and right limits at \( x = 4 \): - Left limit (\( x \to 4^- \)): \( \lim_{x \to 4^-} f(x) = -(4)^2 + 10 = -6 \) - Right limit (\( x \to 4^+ \)): \( \lim_{x \to 4^+} f(x) = 5 - 4 = 1 \) - Actual value at \( x = 4 \): \( f(4) = 1 \) (from the second piece) **Since the left and right limits at \( x=4 \) are NOT equal, the function is NOT continuous at \( x=4 \).** --- ## 4. **Sketch of the Graph** - Draw the downward parabola \( -x^2 + 10 \) from \( x = -4 \) (filled dot at \( (-4, -6) \)) to \( x \to 4 \) (open dot at \( (4, -6) \)). - Draw the line \( 5 - x \) starting at \( (4, 1) \) (filled dot) and continuing to the right. --- ## 5. **Final Answer** ### **Is the function continuous?** **No, the function is not continuous at \( x = 4 \).** --- ## **Summary Table** | Interval | Expression | Endpoint Value(s) | Symbols | |-----------------|------------------|-----------------------------------|-------------| | \([-4, 4)\) | \(-x^2 + 10\) | \((-4, -6)\) (filled), \((4, -6)\) (open) | ●, ○ | | \([4, \infty)\) | \(5 - x\) | \((4, 1)\) (filled) | ● | --- ## **Graph Description** - **Parabola**: Starts at \((-4, -6)\), vertex at \((0, 10)\), ends with open circle at \((4, -6)\). - **Line**: Starts at \((4, 1)\) (filled dot), slopes downward to the right. --- ### **Conclusion** \[ \boxed{ \text{The function is not continuous at } x = 4. } \] Let me know if you need the graph drawn out!

# Graph of the Piecewise Function To visualize the piecewise function defined as: \[ f(x) = \begin{cases} -x^2 + 10 & \text{if } -4 \leq x < 4 \\ 5 - x & \text{if } x \geq 4 \end{cases} \] ## Graph Plotting Steps ### 1. **Plot the First Piece: \( -x^2 + 10 \)** - **Domain**: \( -4 \leq x < 4 \) - **Key Points**: - \( (-4, -6) \) (filled dot) - \( (0, 10) \) (vertex) - \( (4, -6) \) (open dot) ### 2. **Plot the Second Piece: \( 5 - x \)** - **Domain**: \( x \geq 4 \) - **Key Point**: - \( (4, 1) \) (filled dot) ### 3. **Draw the Graph** - Connect the points for \( -x^2 + 10 \) smoothly from \( (-4, -6) \) to \( (4, -6) \) with an open circle at \( (4, -6) \). - From \( (4, 1) \), draw a straight line with a slope of -1 extending to the right. ## Graph Representation ### Graph Sketch ```plaintext 12 | * 11 | 10 | * (0, 10) 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | * (4, 1) 0 |________________________ -4 | -6 | 0 | 4 -6 |* (-4, -6) O (4, -6) -8 | ``` - **Filled dot** at \( (-4, -6) \) and \( (4, 1) \). - **Open dot** at \( (4, -6) \). ## Conclusion The graph visually represents the discontinuity at \( x = 4 \), confirming that the function is not continuous at that point.

# Step-by-Step Solution Let's analyze the corrected piecewise function step by step. --- ## 1. **Understanding the Piecewise Function** The function \( f(x) \) is now defined as: \[ f(x) = \begin{cases} -x^2 + 10 & \text{if } -4 \leq x < 4 \\ -5 - x & \text{if } x \geq 4 \end{cases} \] --- ## 2. **Graph Each Piece** ### **First Piece: \( -x^2 + 10 \) for \( -4 \leq x < 4 \)** - This is a downward-opening parabola. - Domain: \( -4 \leq x < 4 \) - Endpoints: - At \( x = -4 \): \( f(-4) = -(-4)^2 + 10 = -16 + 10 = -6 \) - At \( x = 4 \): \( f(4) = -(4)^2 + 10 = -16 + 10 = -6 \) BUT, since \( x = 4 \) is **not included** (only up to but not including 4), we use an open circle at \( (4, -6) \). - Vertex: \( x=0 \), \( f(0) = -(0)^2 + 10 = 10 \) - Plot points between \( x = -4 \) and \( x = 4 \). ### **Second Piece: \( -5 - x \) for \( x \geq 4 \)** - This is a straight line with slope -1. - Start at \( x = 4 \): \( f(4) = -5 - 4 = -9 \) (included, so solid dot at \( (4, -9) \)). - Continue for \( x > 4 \). --- ## 3. **Check for Continuity at \( x = 4 \)** Let's compare the left and right limits at \( x = 4 \): - Left limit (\( x \to 4^- \)): \[ \lim_{x \to 4^-} f(x) = -(4)^2 + 10 = -6 \] - Right limit (\( x \to 4^+ \)): \[ \lim_{x \to 4^+} f(x) = -5 - 4 = -9 \] - Actual value at \( x = 4 \): \( f(4) = -9 \) (from the second piece). **Since the left and right limits at \( x=4 \) are NOT equal, the function is NOT continuous at \( x=4 \).** --- ## 4. **Sketch of the Graph** - Draw the downward parabola \( -x^2 + 10 \) from \( x = -4 \) (filled dot at \( (-4, -6) \)) to \( x \to 4 \) (open dot at \( (4, -6) \)). - Draw the line \( -5 - x \) starting at \( (4, -9) \) (filled dot) and continuing to the right. --- ## 5. **Final Answer** ### **Is the function continuous?** **No, the function is not continuous at \( x = 4 \).** --- ## **Summary Table** | Interval | Expression | Endpoint Value(s) | Symbols | |-----------------|------------------|-----------------------------------|-------------| | \([-4, 4)\) | \(-x^2 + 10\) | \((-4, -6)\) (filled), \((4, -6)\) (open) | ●, ○ | | \([4, \infty)\) | \(-5 - x\) | \((4, -9)\) (filled) | ● | --- ## **Graph Description** - **Parabola**: Starts at \((-4, -6)\), vertex at \((0, 10)\), ends with open circle at \((4, -6)\). - **Line**: Starts at \((4, -9)\) (filled dot), slopes downward to the right. --- ### **Conclusion** \[ \boxed{ \text{The function is not continuous at } x = 4. } \]