Math Problem Statement

hallar rango y dominio

Solution

To determine the domain, range, and graph of the given piecewise function f(x)f(x), 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 function

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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