Math Problem Statement
Given a piecewise function g(x) defined as follows:
g(x) =
- x^2 - 7 for x < 3
- sqrt(x + 1) for 3 < x < 4
- (2x / (x - 4)) for x ≥ 4
Determine whether g(x) is continuous or not on the following intervals: (-∞, 3), (3, 4), [4, +∞), and (0, 4).
Solution
We are given the function defined piecewise as:
\begin{cases} x^2 - 7, & x < 3 \\ \sqrt{x + 1}, & 3 < x < 4 \\ \frac{2x}{x - 4}, & x \geq 4 \end{cases}$$ We need to determine whether $$g(x)$$ is continuous on the following intervals: 1. $$(-\infty, 3)$$ 2. $$(3, 4)$$ 3. $$[4, +\infty)$$ 4. $$(0, 4)$$ ### Step-by-Step Analysis of Continuity: #### 1. **Interval: $$(-\infty, 3)$$** On this interval, $$g(x) = x^2 - 7$$, a polynomial, which is continuous everywhere. Since the interval does not include 3, there is no issue of discontinuity at the boundary. Thus, $$g(x)$$ is continuous on $$(-\infty, 3)$$. #### 2. **Interval: $$(3, 4)$$** Here, $$g(x) = \sqrt{x + 1}$$. The square root function is continuous as long as the argument is non-negative. For $$x \in (3, 4)$$, $$x + 1$$ is positive, so the function is continuous within this interval. However, $$g(x)$$ is not defined at $$x = 3$$, so the interval excludes 3. Thus, $$g(x)$$ is continuous on $$(3, 4)$$. #### 3. **Interval: $$[4, +\infty)$$** For $$x \geq 4$$, $$g(x) = \frac{2x}{x - 4}$$. This function is undefined at $$x = 4$$ because it causes division by zero. Therefore, $$g(x)$$ has a discontinuity at $$x = 4$$. For $$x > 4$$, the function is continuous since it's a rational function with no further singularities. Thus, $$g(x)$$ is not continuous on $$[4, +\infty)$$. #### 4. **Interval: $$(0, 4)$$** The function needs to be checked across both definitions on this interval: - For $$0 < x < 3$$, $$g(x) = x^2 - 7$$, which is continuous. - For $$3 < x < 4$$, $$g(x) = \sqrt{x + 1}$$, which is also continuous. However, $$g(x)$$ is not defined at $$x = 3$$, and there is no connection between the functions at 3 (since $$g(x)$$ takes on different forms for $$x < 3$$ and $$3 < x$$). Therefore, $$g(x)$$ is discontinuous at $$x = 3$$, so it is not continuous on $$(0, 4)$$. ### Conclusion: 1. $$g(x)$$ is continuous on $$(-\infty, 3)$$. 2. $$g(x)$$ is continuous on $$(3, 4)$$. 3. $$g(x)$$ is **not** continuous on $$[4, +\infty)$$. 4. $$g(x)$$ is **not** continuous on $$(0, 4)$$. Would you like a detailed breakdown of any specific part or have any further questions? Here are five follow-up questions related to this problem: 1. What are the conditions for a function to be continuous at a point? 2. How do we check continuity at a boundary between two function pieces? 3. What types of functions are always continuous within their domains? 4. How can we determine points of discontinuity for rational functions? 5. How would the analysis change if the intervals included 3 or excluded 4? **Tip**: When analyzing piecewise functions, always check for continuity at the points where the definition of the function changes!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Polynomials
Square Root Function
Rational Functions
Formulas
Continuity Check Formula: lim f(x) as x → a = f(a)
Rational Function Formula
Square Root Function
Theorems
Continuity of Polynomials
Continuity of Rational Functions
Discontinuity due to Division by Zero
Suitable Grade Level
Grades 11-12 or early Calculus