Math Problem Statement
i. Sketch and label the graph of the given piecewise function: π(π₯) = { π₯ 3 β 1, π₯ < 0 0, π₯ = 0 βπ₯ + 1 β 2, π₯ > 0 ii. Find the function values at π₯ = β1 and π₯ = 0.
Solution
Solution
(i) Sketch and Label the Graph of the Piecewise Function
The given piecewise function is:
\begin{cases} x^3 - 1, & x < 0 \\ 0, & x = 0 \\ \sqrt{x + 1} - 2, & x > 0 \end{cases}$$ #### **Step 1: Analyze Each Piece** 1. **For $$x < 0$$: $$f(x) = x^3 - 1$$** - This is a cubic function shifted downward by 1. - As $$x \to -\infty$$, $$f(x) \to -\infty$$. - As $$x \to 0^-$$, $$f(x) \to -1$$ (approaching but not including $$x = 0$$). 2. **For $$x = 0$$: $$f(0) = 0$$** - This is a single point at $$(0,0)$$. 3. **For $$x > 0$$: $$f(x) = \sqrt{x + 1} - 2$$** - This is a square root function shifted left by 1 and down by 2. - At $$x = 0$$, $$f(0) = \sqrt{0+1} - 2 = 1 - 2 = -1$$. - As $$x \to \infty$$, $$f(x) \to \infty$$. #### **Step 2: Identify Key Points** - $$f(-1) = (-1)^3 - 1 = -2$$ β Point $$(-1, -2)$$ - $$f(0) = 0$$ β Point $$(0,0)$$ - $$f(1) = \sqrt{1+1} - 2 = \sqrt{2} - 2 \approx -0.59$$ β Point $$(1, -0.59)$$ #### **Step 3: Graphing** - The left branch ($$x^3 - 1$$) is smooth and decreasing. - The function jumps to $$(0,0)$$, making it discontinuous. - The right branch ($$\sqrt{x+1} - 2$$) starts at $$(0,-1)$$ and increases. *(I will generate the graph for visualization.)* numpy as np import matplotlib.pyplot as plt # Define the functions for different pieces def f_left(x): return x**3 - 1 def f_right(x): return np.sqrt(x + 1) - 2 # Define x values for plotting x_left = np.linspace(-2, -0.01, 100) # Left piece (x < 0) x_right = np.linspace(0.01, 2, 100) # Right piece (x > 0) # Compute function values y_left = f_left(x_left) y_right = f_right(x_right) # Plot the functions plt.figure(figsize=(8, 6)) plt.plot(x_left, y_left, 'b', label=r'$f(x) = x^3 - 1$ for $x < 0$') plt.plot(x_right, y_right, 'r', label=r'$f(x) = \sqrt{x+1} - 2$ for $x > 0$') # Plot the discontinuous points plt.scatter(0, 0, color='g', zorder=3, label="Point (0,0)") plt.scatter(0, -1, color='r', edgecolors='black', marker='o', s=50, label="Point (0,-1) (Jump)") # Labels and title plt.xlabel('x') plt.ylabel('f(x)') plt.title('Graph of the Piecewise Function') plt.axhline(0, color='black', linewidth=0.5) plt.axvline(0, color='black', linewidth=0.5) plt.legend() plt.grid(True, linestyle='--', alpha=0.6) plt.show()Here is the graph of the given piecewise function. The key observations: - The left branch ($$x^3 - 1$$) smoothly decreases for $$x < 0$$. - A jump discontinuity occurs at $$x = 0$$, where $$f(0) = 0$$, but the right-hand limit approaches -1. - The right branch ($$\sqrt{x+1} - 2$$) starts at $$(0, -1)$$ and increases. --- ### **(ii) Function Values at $$x = -1$$ and $$x = 0$$** 1. **For $$x = -1$$:** $$f(-1) = (-1)^3 - 1 = -2$$ So, $$f(-1) = -2$$. 2. **For $$x = 0$$:** $$f(0) = 0$$ So, $$f(0) = 0$$. --- ### **Would you like more details or have any questions?** Here are five related questions you can explore: 1. What is the domain and range of this piecewise function? 2. Is the function continuous at $$x = 0$$? Why or why not? 3. What is the left-hand limit and right-hand limit of $$f(x)$$ at $$x = 0$$? 4. How does the graph behave as $$x \to -\infty$$ and $$x \to \infty$$? 5. Can we modify the function to make it continuous at $$x = 0$$? **Tip:** When graphing piecewise functions, always check continuity and discontinuity points carefully.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Graphing
Discontinuity
Cubic Functions
Square Root Functions
Formulas
f(x) = x^3 - 1 for x < 0
f(x) = β(x + 1) - 2 for x > 0
f(x) = 0 for x = 0
Theorems
Continuity of piecewise functions
Limits for piecewise functions
Suitable Grade Level
Grades 9-11
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