Problem 31 - Precalculus with Calculus Previews

Question

In Problems \(25-32\), find the domain of the given function \(f .\) Find the \(x\) -intercept and the vertical asymptote of the graph. Use transformations to graph the given function \(f\). $$ f(x)=-1+\ln x $$

Step-by-Step Solution

Answer

Domain: \((0, \infty)\), x-intercept: \((e, 0)\), vertical asymptote: \(x = 0\).

Step 1: Determine the Domain

The function given is \( f(x) = -1 + \ln x \). The natural logarithm function \( \ln x \) is only defined for positive numbers. This implies that \( x > 0 \) for \( \ln x \) to be valid. Hence, the domain of \( f(x) \) is all positive real numbers, \((0, \infty)\).

Step 2: Find the x-intercept

To find the x-intercept, we set \( f(x) = 0 \) and solve for \( x \): \[ -1 + \ln x = 0 \] \[ \ln x = 1 \] \[ x = e \] where \( e \) is the base of the natural logarithm, approximately 2.718. Thus, the \( x \)-intercept is \((e, 0)\).

Step 3: Find the Vertical Asymptote

A vertical asymptote occurs where the function is undefined or where the function tends to infinity as \( x \) approaches a certain value. Since \( \ln x \) approaches negative infinity as \( x \to 0^+ \), the vertical asymptote for \( f(x) = -1 + \ln x \) is at \( x = 0 \).

Step 4: Use Transformations to Graph the Function

The function \( f(x) = -1 + \ln x \) can be graphed by starting with the graph of \( \ln x \) and then shifting it downward by 1 unit. The graph originally has a vertical asymptote at \( x = 0 \), and the shift does not affect this asymptote. The function passes through the \( x \)-intercept \((e, 0)\) after the transformation.

Key Concepts

X-Intercept

The x-intercept of a function is the point where the graph crosses the x-axis. This means it is the point where the output value (y) is zero. To find the x-intercept, we set the function equal to zero and solve for x.
For the function given, \[ f(x) = -1 + \ln x \],we find the x-intercept by solving: \[ -1 + \ln x = 0 \],which simplifies to:

• \( \ln x = 1 \)
• \( x = e \)

The solution \( x = e \) tells us that the function intersects the x-axis at this point. Remember, \( e \) is approximately 2.718, which is the base of the natural logarithm. Thus, the x-intercept of this function is at the point \((e, 0)\). Knowing the x-intercept is essential for sketching the graph accurately, as it gives a point of reference where the function changes sign.

Vertical Asymptote

A vertical asymptote occurs where the function approaches infinity and is undefined. For logarithmic functions such as \( \ln x \), vertical asymptotes typically occur where the argument of the logarithm is zero because the natural logarithm is undefined at and does not cross zero.
In our function \( f(x) = -1 + \ln x \), the vertical asymptote is determined by considering where \( \ln x \) becomes undefined. Since \( \ln x \) tends toward negative infinity as \( x \to 0^+ \), we have a vertical asymptote at \( x = 0 \). This means the graph will approach the line \( x = 0 \) but will never touch or cross it.
Vertical asymptotes are crucial aspects of sketching graphs because they define boundaries beyond which the function cannot exist. They help to visualize the behavior of the function near these boundary points.

Graph Transformations

Graph transformations are modifications applied to the base graph of a function to simplify the process of sketching visually. Transformations include shifts, stretches, and reflections among others.
For the function \( f(x) = -1 + \ln x \), the transformation involves shifting the graph of the standard \( \ln x \) function.

• The "-1" in the function indicates a vertical shift downward by 1 unit.

When graphing,

• Start by sketching the basic graph of \( \ln x \), which passes through the point \((1, 0)\), has a vertical asymptote at \( x = 0 \), and increases slowly without bounds.
• Then, shift every point on this graph 1 unit down to get the final graph of \( f(x) = -1 + \ln x \).

The vertical asymptote remains at \( x = 0 \) because the downward shift does not affect this feature. It is helpful to note that such transformations simplify how complex graphs can be understood and explained.