Determine Continuity of a Piecewise Rational Function with Discontinuities at x = 2 and x = -2

Math Problem Statement

At what points is the following function continuous?

f(x)equals=left brace Start 3 By 2 Matrix 1st Row 1st Column StartFraction x cubed plus 8 Over x squared minus 4 EndFraction comma 2nd Column x not equals 2 comma x not equals minus 2 2nd Row 1st Column negative 3 comma 2nd Column x equals negative 2 3rd Row 1st Column 8 comma 2nd Column x equals 2 EndMatrix

x3+8x2−4,

x≠2, x≠−2

−3,

x=−2

x=2

Question content area bottom

Part 1

Choose the correct answer below and, if necessary, fill in the answer box to complete your choice.

All real numbers except at

xequals=enter your response here

(Use a comma to separate answers as needed.)

Only at

xequals=enter your response here

(Use a comma to separate answers as needed.)

All real numbers x

Solution

We are given a piecewise function, and we need to determine at which points this function is continuous. Let’s break it down.

The function $f(x)$ is defined as follows:

\begin{cases} \frac{x^3 + 8}{x^2 - 4}, & \text{if } x \neq 2 \text{ and } x \neq -2 \\ -3, & \text{if } x = -2 \\ 8, & \text{if } x = 2 \end{cases}$$ ### Step 1: Analyze $$f(x)$$ when $$x \neq 2$$ and $$x \neq -2$$ The function is given by: $$f(x) = \frac{x^3 + 8}{x^2 - 4} = \frac{x^3 + 8}{(x-2)(x+2)}$$ This rational function has a potential issue at the points where the denominator is zero, which happens when $$x = 2$$ or $$x = -2$$. Thus, $$f(x)$$ has **discontinuities at $$x = 2$$ and $$x = -2$$** for now. ### Step 2: Check the behavior at $$x = -2$$ At $$x = -2$$, the function value is explicitly given as $$f(-2) = -3$$. To check for continuity, we need the limit as $$x$$ approaches $$-2$$ from both sides. First, simplify $$f(x)$$: $$f(x) = \frac{(x + 2)(x^2 - 2x + 4)}{(x-2)(x+2)} = \frac{x^2 - 2x + 4}{x - 2} \text{ for } x \neq 2 \text{ and } x \neq -2$$ Now, evaluate the limit as $$x \to -2$$: $$\lim_{x \to -2} f(x) = \lim_{x \to -2} \frac{x^2 - 2x + 4}{x - 2}$$ Substitute $$x = -2$$: $$\lim_{x \to -2} f(x) = \frac{(-2)^2 - 2(-2) + 4}{-2 - 2} = \frac{4 + 4 + 4}{-4} = -3$$ Since $$f(-2) = -3$$, the function is continuous at $$x = -2$$. ### Step 3: Check the behavior at $$x = 2$$ At $$x = 2$$, the function value is explicitly given as $$f(2) = 8$$. Now, we check the limit as $$x \to 2$$: $$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^2 - 2x + 4}{x - 2}$$ Substituting $$x = 2$$ gives an undefined form, but we can evaluate the limit algebraically. Notice that the function cannot be simplified further to avoid the discontinuity at $$x = 2$$, meaning $$\lim_{x \to 2} f(x)$$ does not exist in a way that equals $$8$$. Therefore, **the function is discontinuous at $$x = 2$$**. ### Conclusion: - $$f(x)$$ is continuous at all points except $$x = 2$$ and $$x = -2$$. Thus, the correct answer is: $$\boxed{A. \text{All real numbers except at } x = 2, -2}$$ --- Would you like further details or have any questions? ### Related questions: 1. How do you prove that a function is continuous at a given point? 2. What are the conditions for a rational function to be continuous? 3. How can we use limits to determine discontinuity? 4. What happens at points where the denominator of a function is zero? 5. How do piecewise functions behave at the boundaries of their definitions? ### Tip: When dealing with rational functions, always check for points where the denominator is zero, as these are potential discontinuities.

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Math Problem Analysis

Mathematical Concepts

Continuity
Piecewise Functions
Rational Functions
Limits

Formulas

f(x) = (x^3 + 8)/(x^2 - 4)
Limit of rational functions
Continuity at a point

Theorems

Continuity Theorem
Limit Laws
Piecewise Function Continuity

Suitable Grade Level

Grades 11-12 (Calculus introduction)

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