Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ f(x)=\ln \left(x^{2}+100\right) $$. Find the domain. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain is all real $$ x $$, except $$ x= $$ (Use a comma to separate answers as needed.) Bhe domain is all real numbers.

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**Step 1. Determine the domain of $$ f(x)=\ln\left(x^2+100\right) $$:**  
The logarithm function is defined only for positive arguments. Thus, we require  
$$
x^2+100>0.
$$  
Since $$ x^2 \geq 0 $$ for all real $$ x $$, it follows that  
$$
x^2+100\geq100>0 \quad \text{for all } x\in\mathbb{R}.
$$  
Thus, the expression inside the logarithm is always positive.

**Conclusion for the Domain:**  
The domain of $$ f(x) $$ is all real numbers. This corresponds to answer choice B.

---

**Step 2. Summarize Pertinent Graph Information by Applying the Graphing Strategy:**

1. **Domain:**  
   $$
   x\in\mathbb{R}.
   $$

2. **Range:**  
   Notice that the smallest value of $$ x^2 $$ is 0, attained when $$ x=0 $$. Therefore,  
   $$
   f(0)=\ln(0^2+100)=\ln(100).
   $$  
   As $$ |x| $$ increases, $$ x^2+100 $$ increases without bound, so $$ \lim_{|x|\to\infty}\ln\left(x^2+100\right)=+\infty $$. Hence, the range is  
   $$
   \left[\ln(100),\infty\right).
   $$

3. **Symmetry:**  
   The function is even because $$ x^2 $$ is even. Thus,  
   $$
   \ln\left(x^2+100\right)=\ln\left((-x)^2+100\right),
   $$  
   meaning the graph is symmetric about the $$ y $$-axis.

4. **Intercepts:**  
   - **$$ y $$-intercept:** Set $$ x=0 $$ to obtain  
     $$
     f(0)=\ln(100).
     $$
   - **$$ x $$-intercepts:** To find $$ x $$-intercepts, solve $$ \ln\left(x^2+100\right)=0 $$.  
     Exponentiating both sides gives:
     $$
     x^2+100=1 \quad \Longrightarrow \quad x^2=-99.
     $$
     This has no real solution, so there are no $$ x $$-intercepts.

5. **End Behavior:**  
   As $$ |x| \to\infty $$,  
   $$
   f(x)=\ln\left(x^2+100\right) \sim \ln\left(x^2\right)=2\ln|x|,
   $$  
   which increases without bound, though more slowly than any polynomial.

---

**Step 3. Sketch the Graph of $$ f(x)=\ln\left(x^2+100\right) $$:**

- **Symmetry:** The graph is symmetric about the $$ y $$-axis.  
- **Minimum Value:** The lowest point occurs at $$ x=0 $$ with $$ f(0)=\ln(100) $$.  
- **Increasing Behavior:** The function increases as $$ |x| $$ increases, approaching $$ +\infty $$ as $$ x \to \pm\infty $$.  
- **No Asymptotes:** Since the argument inside the logarithm is always greater than zero, there are no vertical asymptotes. The graph has a horizontal asymptotic behavior as $$ x \to \pm\infty $$ in the sense of slowly unbounded increase.

---

**Final Answer:**  
B. The domain is all real numbers.
The domain of $$ f(x) = \ln(x^2 + 100) $$ is all real numbers.

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