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 , we need to verify the following three conditions:
- The function is defined at .
- The limit of the function as approaches exists.
- The limit of the function as approaches is equal to the function's value at .
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.Ask a new question for Free
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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