Math Problem Statement
hallar rango y dominio
Solution
To determine the domain, range, and graph of the given piecewise function , let's analyze each part of the function and its domain separately.
The function is given as:
(x+2)^2 + 1 & \text{if } x \in (-\infty, -4] \\ \sqrt{x+1} & \text{if } x \in (0, 4) \\ 2x - 8 & \text{if } x \in [4, \infty) \end{cases}$$ ### Domain The domain of $$f(x)$$ is the set of all $$x$$ values for which $$f(x)$$ is defined. Each piece of the function has its own domain: 1. $$(x+2)^2 + 1$$ is defined for $$x \in (-\infty, -4]$$. 2. $$\sqrt{x+1}$$ is defined for $$x \in (0, 4)$$ (as the square root function requires $$x+1 \geq 0$$). 3. $$2x - 8$$ is defined for $$x \in [4, \infty)$$. Combining these, the domain of $$f(x)$$ is: $$\text{Domain} = (-\infty, -4] \cup (0, 4) \cup [4, \infty)$$ ### Range Next, we find the range of each piece of the function: 1. For $$(x+2)^2 + 1$$ when $$x \in (-\infty, -4]$$: - As $$x$$ approaches $$-\infty$$, $$(x+2)^2$$ approaches $$\infty$$, so $$(x+2)^2 + 1$$ also approaches $$\infty$$. - At $$x = -4$$, $$f(x) = (-4+2)^2 + 1 = 4 + 1 = 5$$. - Therefore, the range for this part is $$[5, \infty)$$. 2. For $$\sqrt{x+1}$$ when $$x \in (0, 4)$$: - When $$x = 0^+$$, $$\sqrt{x+1}$$ approaches $$\sqrt{1} = 1$$. - When $$x = 4^-$$, $$\sqrt{x+1}$$ approaches $$\sqrt{5}$$. - Therefore, the range for this part is $$(1, \sqrt{5})$$. 3. For $$2x - 8$$ when $$x \in [4, \infty)$$: - When $$x = 4$$, $$2x - 8 = 0$$. - As $$x$$ approaches $$\infty$$, $$2x - 8$$ also approaches $$\infty$$. - Therefore, the range for this part is $$[0, \infty)$$. Combining these, the range of $$f(x)$$ is: $$\text{Range} = [5, \infty) \cup (1, \sqrt{5}) \cup [0, \infty) = [0, \infty)$$ ### Graph I'll now plot the graph of the function $$f(x)$$ based on the defined pieces. Let's proceed to visualize it using Python. ```python import matplotlib.pyplot as plt import numpy as np # Define the pieces of the function x1 = np.linspace(-10, -4, 400) x2 = np.linspace(0, 4, 400) x3 = np.linspace(4, 10, 400) y1 = (x1 + 2)**2 + 1 y2 = np.sqrt(x2 + 1) y3 = 2*x3 - 8 # Plot the function plt.figure(figsize=(10, 6)) plt.plot(x1, y1, label=r'$(x+2)^2 + 1, x \in (-\infty, -4]$') plt.plot(x2, y2, label=r'$\sqrt{x+1}, x \in (0, 4)$') plt.plot(x3, y3, label=r'$2x - 8, x \in [4, \infty)$') # Highlight the boundary points plt.scatter([-4], [(4+2)**2 + 1], color='red') # point at x = -4 plt.scatter([0], [np.sqrt(0+1)], color='red') # point at x = 0 plt.scatter([4], [np.sqrt(4+1)], color='red') # point at x = 4, sqrt functionAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Domain and Range
Square Roots
Linear Functions
Formulas
f(x) = (x+2)^2 + 1, f(x) = sqrt(x+1), f(x) = 2x - 8
Theorems
Properties of piecewise functions
Domain and range determination
Suitable Grade Level
Grades 10-12