To analyze the function $y = \log_2(x^2 + 1)$, let's delve into several key aspects:

1. Domain: $\text{The domain of the function is the set of all } x \text{ for which the argument of the logarithm is positive.}$ $x^2 + 1 > 0 \quad \text{for all real numbers } x$ $\text{Thus, the domain is } (-\infty, \infty).$

2. Range: $\text{The range of the function is determined by the values that } y \text{ can take.}$ $x^2 \geq 0 \implies x^2 + 1 \geq 1$ $\log_2(x^2 + 1) \geq \log_2(1) = 0$ $\text{As } x^2 \to \infty, \; x^2 + 1 \to \infty \implies \log_2(x^2 + 1) \to \infty$ $\text{Thus, the range is } [0, \infty).$

3. Behavior and Symmetry: $\text{Since } x^2 \text{ is an even function, } x^2 + 1 \text{ is also an even function.}$ $\log_2(x^2 + 1) \text{ is also even, which means it is symmetric about the y-axis.}$ $\log_2(-x^2 + 1) = \log_2(x^2 + 1)$

4. Intercepts:

   • y-intercept: $\text{Set } x = 0 \implies y = \log_2(0^2 + 1) = \log_2(1) = 0$ $\text{The y-intercept is } (0, 0).$
   • x-intercepts: $\text{There are no x-intercepts because } \log_2(x^2 + 1) \geq 0 \text{ for all } x.$

5. Asymptotes: $\text{There are no vertical asymptotes since the logarithm's argument is always positive.}$ $\text{There are no horizontal asymptotes since } \log_2(x^2 + 1) \text{ increases without bound as } x \to \infty.$

6. First Derivative (to find critical points and analyze increasing/decreasing behavior): $y' = \frac{d}{dx} \log_2(x^2 + 1)$ $y' = \frac{1}{\ln(2)} \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{(x^2 + 1) \ln(2)}$ $\text{Setting } y' = 0 \text{ gives } x = 0.$ $\text{For } x > 0, y' > 0 \text{ and for } x < 0, y' < 0.$ $\text{Thus, the function is increasing for } x > 0 \text{ and decreasing for } x < 0.$

7. Second Derivative (to determine concavity): $y'' = \frac{d}{dx} \left( \frac{2x}{(x^2 + 1) \ln(2)} \right)$ $y'' = \frac{2((x^2 + 1) \ln(2) - 2x^2)}{((x^2 + 1) \ln(2))^2} = \frac{2(\ln(2) - x^2 \ln(2))}{((x^2 + 1) \ln(2))^2}$ $y'' = \frac{2(1 - x^2)}{(x^2 + 1)^2 \ln(2)}$ $\text{Setting } y'' = 0 \text{ gives } x^2 = 1 \implies x = \pm 1.$

   $\text{For } |x| < 1, y'' > 0 \implies \text{concave up}.$ $\text{For } |x| > 1, y'' < 0 \implies \text{concave down}.$

Would you like any additional details or have any questions about this analysis?

Here are 8 related questions to expand on this information:

1. How do you find the domain of a logarithmic function?
2. What are the properties of logarithms?
3. Can you explain how to determine the range of a function?
4. How do you find the intercepts of a function?
5. What are the steps to find the first derivative of a function?
6. How does the first derivative test help in finding local maxima and minima?
7. What does the second derivative tell us about the concavity of a function?
8. How can you sketch the graph of a logarithmic function given its key characteristics?

Tip: Understanding the properties of logarithms can greatly simplify solving and analyzing logarithmic functions.