24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.
ƒ(x) = ln( x² + 3) / (x -1)

Verified step by step guidance

1. Identify the domain of the function \( f(x) = \frac{\ln(x^2 + 3)}{x - 1} \). The natural logarithm function \( \ln(x^2 + 3) \) is defined for all real numbers since \( x^2 + 3 > 0 \) for all \( x \). However, the denominator \( x - 1 \) cannot be zero, so \( x \neq 1 \). Thus, the domain is all real numbers except \( x = 1 \).
2. Find the vertical asymptote by setting the denominator equal to zero: \( x - 1 = 0 \). This gives \( x = 1 \) as a vertical asymptote.
3. Determine the behavior of the function as \( x \to \infty \) and \( x \to -\infty \). As \( x \to \infty \), \( \ln(x^2 + 3) \approx \ln(x^2) = 2\ln(x) \), and the function behaves like \( \frac{2\ln(x)}{x} \), which approaches 0. Similarly, as \( x \to -\infty \), the function also approaches 0.
4. Find the first derivative \( f'(x) \) to determine critical points and intervals of increase or decrease. Use the quotient rule: \( f'(x) = \frac{(x - 1) \cdot \frac{d}{dx}[\ln(x^2 + 3)] - \ln(x^2 + 3) \cdot \frac{d}{dx}[x - 1]}{(x - 1)^2} \). Simplify to find critical points.
5. Find the second derivative \( f''(x) \) to determine concavity and points of inflection. Use the derivative of \( f'(x) \) and analyze the sign of \( f''(x) \) to determine where the function is concave up or down.

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function f(x) = ln(x² + 3) / (x - 1), the domain excludes x = 1, where the denominator becomes zero, causing a division by zero error. Additionally, the expression inside the logarithm, x² + 3, must be positive, which it always is for real numbers.

Asymptotes

Asymptotes are lines that a graph approaches but never touches. For f(x) = ln(x² + 3) / (x - 1), there is a vertical asymptote at x = 1 due to the division by zero. Horizontal asymptotes can be determined by analyzing the behavior of the function as x approaches infinity or negative infinity, which involves understanding the limits of the function.

Critical Points and Derivatives

Critical points occur where the derivative of a function is zero or undefined, indicating potential maxima, minima, or inflection points. To find these for f(x) = ln(x² + 3) / (x - 1), compute the first derivative and solve for x where the derivative equals zero or is undefined. This helps in understanding the function's increasing or decreasing behavior and concavity.