Continuous Piecewise Function Problem with Limit Analysis at x = 3

Math Problem Statement

Consider two functions, 𝑓: 𝐷𝑓 ⊂ ℝ → ℝ and 𝑔: ℝ → ℝ, such that:

 𝑓 is defined by 𝑓(𝑥)

 𝑔 is defined by 𝑔(𝑥) = { 𝑓(𝑥) if 𝑥 < 3

ln(𝑥2 − 8) if 𝑥 ≥ 3

 𝑔 is continuous

Which of the following may be 𝑓(𝑥)?

A 𝑒𝑥−3

𝑥2 − 9

𝑥 − 3

sin(𝑥 − 3)

− 1

𝑥 − 3

D 3

Solution

We are given two functions, $f$ and $g$ , with $f$ defined on some subset of $\mathbb{R}$ , and $g$ defined piecewise as follows:

\begin{cases} f(x), & \text{if } x < 3 \\ \ln(x^2 - 8), & \text{if } x \geq 3 \end{cases}$$ We are also told that $$g(x)$$ is continuous on $$\mathbb{R}$$. To determine which of the given options could be $$f(x)$$, we need to ensure the continuity of $$g(x)$$ at $$x = 3$$. ### Step 1: Analyze the continuity condition at $$x = 3$$ For $$g(x)$$ to be continuous at $$x = 3$$, the left-hand limit of $$g(x)$$ as $$x \to 3^{-}$$ must equal the right-hand limit as $$x \to 3^{+}$$, which must also equal the value of $$g(3)$$. - **Right-hand limit**: As $$x \to 3^{+}$$, we use the second branch of the piecewise definition, $$g(x) = \ln(x^2 - 8)$$. Therefore, we calculate: $$\lim_{x \to 3^{+}} g(x) = \ln(3^2 - 8) = \ln(9 - 8) = \ln(1) = 0$$ Hence, $$\lim_{x \to 3^{+}} g(x) = 0$$. - **Left-hand limit**: As $$x \to 3^{-}$$, we use the first branch of $$g(x)$$, which is $$f(x)$$. Therefore, for continuity, we require: $$\lim_{x \to 3^{-}} f(x) = 0$$ This means $$f(x)$$ must approach 0 as $$x \to 3^{-}$$. ### Step 2: Analyze the options for $$f(x)$$ We now evaluate each of the given functions $$f(x)$$ to determine which satisfies $$\lim_{x \to 3^{-}} f(x) = 0$$. - **Option A: $$f(x) = e^{x - 3}$$** As $$x \to 3^{-}$$, $$\lim_{x \to 3^{-}} e^{x - 3} = e^{3 - 3} = e^0 = 1$$ This does not satisfy the required condition $$\lim_{x \to 3^{-}} f(x) = 0$$, so **Option A is incorrect**. - **Option B: $$f(x) = \frac{x^2 - 9}{x - 3}$$** We simplify this expression: $$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3, \quad \text{for } x \neq 3$$ As $$x \to 3^{-}$$, $$\lim_{x \to 3^{-}} (x + 3) = 6$$ This does not satisfy $$\lim_{x \to 3^{-}} f(x) = 0$$, so **Option B is incorrect**. - **Option C: $$f(x) = \frac{\sin(x - 3)}{x - 3}$$** As $$x \to 3^{-}$$, this is an indeterminate form $$\frac{0}{0}$$. Using L'Hopital's Rule, we compute the limit: $$\lim_{x \to 3^{-}} \frac{\sin(x - 3)}{x - 3} = \lim_{x \to 3^{-}} \frac{\cos(x - 3)}{1} = \cos(0) = 1$$ This does not satisfy $$\lim_{x \to 3^{-}} f(x) = 0$$, so **Option C is incorrect**. - **Option D: $$f(x) = 3$$** As $$x \to 3^{-}$$, $$\lim_{x \to 3^{-}} f(x) = 3$$ This does not satisfy $$\lim_{x \to 3^{-}} f(x) = 0$$, so **Option D is incorrect**. ### Conclusion: None of the provided options satisfy the condition that $$\lim_{x \to 3^{-}} f(x) = 0$$. Therefore, there is **no valid option** from the list that could be $$f(x)$$. This problem might involve an

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Continuity of Functions
Piecewise Functions
Limits
L'Hopital's Rule

Formulas

Limit formula: lim(x → c) f(x)
L'Hopital's Rule for indeterminate forms

Theorems

Continuity Theorem
L'Hopital's Rule

Suitable Grade Level

Grades 11-12, University Level

Related Recommendation

Determine Continuity of Piecewise Function g(x) at x = 0

Analyzing Continuity and Limits of a Piecewise Function at x = 3

Identifying a Function Based on Limits and Continuity

Step-by-Step Analysis of Piecewise Function with Limits and Continuity

Evaluate Limits of a Piecewise Function Involving x, x^2, and 8-x