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Understanding the domain and range of a logarithmic function is crucial in grasping the characteristics of the function. Let's break it down: The function given, \( f(x) = \ln x + 2 \), has a natural logarithm component, \( \ln x \). Logarithms are only defined for positive numbers. Thus, the domain of \( f(x) \) is all positive real numbers, which means that \( x > 0 \). In simpler terms, any input for \( x \) must be greater than zero.

The range of a function tells us what output values we can get. For the natural logarithm, \( \ln x \) its range spans all real numbers, from \(-\infty\) to \( \infty \). Adding a constant like 2 shifts the graph vertically upwards but does not alter the span of potential outcomes. Therefore, the range of \( f(x) = \ln x + 2 \) remains the set of all real numbers, \( (-\infty, \infty) \).

• Domain of \( f(x) \): \( x > 0 \)
• Range of \( f(x) \): \( (-\infty, \infty) \)

The x-intercept of a function graphically represents where the curve crosses the x-axis. This intersection is a point where the function equals zero. To find it for \( f(x) = \ln x + 2 \), we set the function to zero:

• \( \ln x + 2 = 0 \)
• Simplify to \( \ln x = -2 \)
• Solve for \( x \) using the property of logarithms: \( x = e^{-2} \)

The x-intercept occurs at the point \( (e^{-2}, 0) \). This is a specific spot on the graph, where the height, or \( y \)-value, is zero.

Understanding how to calculate x-intercepts helps establish important points on your graph and gain insights into the behavior of the function near that point.

Graph sketching provides a visual representation of a function, helping to analyze its behavior and how it changes across its domain. For \( f(x) = \ln x + 2 \), we can use insights from the parent function \( \ln x \), which is a well-known curve that passes through the point (1, 0) and steadily increases.

Here's how the graph of \( f(x) = \ln x + 2 \) can be visualized:

• Start with the graph of \( \ln x \), which naturally starts near the origin but on the positive side of the x-axis.
• Shift the entire graph upwards by 2 units. This shift represents the \(+2\) in \( \ln x + 2 \), adjusting all y-values by 2.
• Mark the x-intercept \( (e^{-2}, 0) \), as identified earlier. This point is crucial for sketching.
• Draw the curve so it starts near the x-axis for large negative values of \( y \) (indicating increasing values of \( x \)), and rises steadily to the right as \( x \) increases.

These steps offer a simple method to sketch the function, providing clarity and aiding in further analysis.