Math Problem Statement

Show that the following functions are continuous at the specified point: (a) f(x) = (x^2 - 16)/(x - 4) if x ≠ 4, 8 if x = 4, with a = 4 (b) f(x) = 2 + 3x + 4x^2 if x ≤ 1, 4 + 3x + 2x^2 if x > 1, with a = 1 (c) f(x) = 1 - x^2 if x ≤ 1, 1 - x^3 if x > 1, with a = 1

Solution

To show that each function is continuous at the specified point aa, we need to verify the following three conditions:

  1. The function is defined at aa.
  2. The limit of the function as xx approaches aa exists.
  3. The limit of the function as xx approaches aa is equal to the function's value at aa.

Let's analyze each part:

Part (a)

Given:

\frac{x^2 - 16}{x - 4} & \text{if } x \neq 4 \\ 8 & \text{if } x = 4 \end{cases} \quad \text{and} \quad a = 4$$ 1. **Function value at $$a = 4$$:** $$f(4) = 8$$ 2. **Limit as $$x$$ approaches 4:** The expression for $$f(x)$$ when $$x \neq 4$$ is: $$\frac{x^2 - 16}{x - 4} = \frac{(x - 4)(x + 4)}{x - 4} = x + 4$$ Therefore, $$\lim_{x \to 4} f(x) = \lim_{x \to 4} (x + 4) = 4 + 4 = 8$$ 3. **Check continuity:** Since $$\lim_{x \to 4} f(x) = f(4) = 8$$, $$f(x)$$ is continuous at $$x = 4$$. ### Part (b) Given: $$f(x) = \begin{cases} 2 + 3x + 4x^2 & \text{if } x \leq 1 \\ 4 + 3x + 2x^2 & \text{if } x > 1 \end{cases} \quad \text{and} \quad a = 1$$ 1. **Function value at $$a = 1$$:** $$f(1) = 2 + 3(1) + 4(1)^2 = 2 + 3 + 4 = 9$$ 2. **Limit as $$x$$ approaches 1 from the left ($$x \to 1^-$$):** $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(2 + 3x + 4x^2\right) = 2 + 3(1) + 4(1)^2 = 9$$ 3. **Limit as $$x$$ approaches 1 from the right ($$x \to 1^+$$):** $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(4 + 3x + 2x^2\right) = 4 + 3(1) + 2(1)^2 = 9$$ 4. **Check continuity:** Since both the left-hand limit and right-hand limit are equal, and they match the function value $$f(1) = 9$$, $$f(x)$$ is continuous at $$x = 1$$. ### Part (c) Given: $$f(x) = \begin{cases} 1 - x^2 & \text{if } x \leq 1 \\ 1 - x^3 & \text{if } x > 1 \end{cases} \quad \text{and} \quad a = 1$$ 1. **Function value at $$a = 1$$:** $$f(1) = 1 - 1^2 = 1 - 1 = 0$$ 2. **Limit as $$x$$ approaches 1 from the left ($$x \to 1^-$$):** $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(1 - x^2\right) = 1 - 1^2 = 0$$ 3. **Limit as $$x$$ approaches 1 from the right ($$x \to 1^+$$):** $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(1 - x^3\right) = 1 - 1^3 = 0$$ 4. **Check continuity:** Since both the left-hand limit and right-hand limit are equal, and they match the function value $$f(1) = 0$$, $$f(x)$$ is continuous at $$x = 1$$. ### Summary Each function $$f(x)$$ is continuous at the specified point $$a$$ after verifying that the function's limit as $$x$$ approaches $$a$$ exists and equals the function value at $$a$$. Would you like to explore any of these in more detail or have any questions? --- Here are 5 related questions you might consider: 1. How would you determine if a function is differentiable at a point $$a$$? 2. What happens if the left-hand and right-hand limits at $$a$$ are not equal? 3. Can a function be continuous at a point but not differentiable there? Provide an example. 4. How does the continuity of piecewise functions differ from standard functions? 5. What are the implications of a function being continuous on a closed interval? **Tip:** When dealing with piecewise functions, always evaluate the limits from both sides of the point in question to ensure continuity.

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Math Problem Analysis

Mathematical Concepts

Continuity
Piecewise Functions
Limits

Formulas

Limit Definition: lim(x -> a) f(x) = L
Quadratic Expansion: (x^2 - 16) = (x - 4)(x + 4)

Theorems

Continuity Theorem: A function f(x) is continuous at a point a if f(a) is defined, lim(x -> a) f(x) exists, and lim(x -> a) f(x) = f(a)

Suitable Grade Level

Grade 12 or Early College