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

B

𝑥2 − 9

𝑥 − 3

C

sin(𝑥 − 3)

− 1

𝑥 − 3

D 3

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

\[
g(x) = 
\begin{cases} 
f(x), & \text{if } x < 3 \\
\ln(x^2 - 8), & \text{if } x \geq 3
\end{cases}
\]
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

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, and 𝑔 is continuous. Which of the following may be 𝑓(𝑥)? A) 𝑒𝑥−3, B) (𝑥2 − 9)/(𝑥 − 3), C) sin(𝑥 − 3)/(𝑥 − 3), D) 3

Explore a piecewise function problem where g(x) is continuous, and f(x) must satisfy specific conditions at x = 3. Analyze limit behavior using continuity conditions and L'Hopital's Rule. This problem covers advanced topics suitable for Grades 11-12 and early university levels.

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