Problem 74 - College Algebra and Calculus: An Applied Approach

Question

Find the domain, vertical asymptote, and \(x\) -intercept of the logarithmic function. Then sketch its graph.\(f(x)=3+\ln x\)

Step-by-Step Solution

Answer

The domain of the function \(f(x) = 3 + \ln x\) is (0, +∞). The x-intercept is \(x = e^{-3}\) and the vertical asymptote is x = 0. The graph of the function is a graph of f(x) = ln(x), shifted upward by 3 units.

Step 1: Finding the Domain

The first step is to determine the domain of the logarithm function \(f(x) = 3 + \ln x\). The argument of the natural logarithm function must always be greater than zero. So, x > 0. Therefore, the domain of the function is (0, +∞).

Step 2: Finding the x - intercept

The x-intercept is the point where the function crosses the x-axis, in other words, where \(f(x) = 0\). To find this point, set the function equal to zero and solve for x: \(0 = 3 + \ln x \Rightarrow -3 = \ln x\). Exponentiate both sides using base e: \(e^{-3} = x \approx 0.0498\).

Step 3: Find the Vertical Asymptotes

The vertical asymptote for a logarithmic function like \(f(x) = 3 + \ln x\) is \(x = 0\). This is a vertical line on the x-axis this function will approach but never touch.

Step 4: Sketching the Graph

When sketching logarithmic functions, it is useful to remember the basic shape of natural logarithm function, then shifting it as per this function. In this case, sketch the graph of f(x) = ln(x), but shift it upward by 3 units. Ahead of x=0, the graph will approach the vertical asymptote, with the x-intercept at \(x = e^{-3}\) and the y-intercept at \(y = 3\).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial, as it tells us the set of all possible input values for which the function is defined. For a logarithmic function like \(f(x) = 3 + \ln x\), the domain is determined by the requirement that the argument of the logarithm (in this case, \(x\)) must be greater than zero.
This is due to the fact that the natural logarithm is undefined for non-positive values. Therefore, the domain of the function \(f(x) = 3 + \ln x\) is all positive real numbers, which can be expressed in interval notation as \( (0, +∞)\).
This means that you can substitute any positive value for \(x\), and the function will output a real number. So, always remember: when dealing with logarithmic functions, check that the input is greater than zero! This simple yet important idea of domain helps us avoid errors when evaluating or sketching the function.

Vertical Asymptote

A vertical asymptote is a vertical line that a graph approaches but never actually touches. It often indicates where a function takes on values that grow infinitely large in magnitude.
For the function \(f(x) = 3 + \ln x\), there is a vertical asymptote at \(x = 0\). This happens because the natural logarithm \(\ln x\) becomes undefined as \(x\) approaches zero from the right, causing the values of the function to tend towards negative infinity.
When sketching the graph of the function, you will see that, as \(x\) gets closer to zero, the graph of \(f(x)\) sharply drops downwards, illustrating that the function "shoots down" towards \(-∞\). Vertical asymptotes are key to sketching functions accurately because they help us understand the behavior of the graph near certain critical values.

X-Intercept

The \(x\)-intercept is the point where the graph of a function crosses the \(x\)-axis. For \(f(x) = 3 + \ln x\), you find this point by setting the function equal to zero and solving for \(x\). This means finding the value of \(x\) such that \(f(x) = 0\).
After setting the equation \(0 = 3 + \ln x\), the next step is solving for \(x\) which gives \(-3 = \ln x\). To go further, exponentiate both sides with base \(e\) to isolate \(x\). Doing this, we find \(x = e^{-3}\), which approximates to \(0.0498\).
Hence, the \(x\)-intercept of this function is at the point \((e^{-3}, 0)\). This point is an essential part of sketching the graph, as it tells us exactly where the curve will cross the \(x\)-axis.

Graph Sketching

Sketching the graph of a logarithmic function may seem daunting at first, but understanding its key features simplifies the process. To sketch \(f(x) = 3 + \ln x\), start with recognizing that the graph of a basic natural logarithm \(\ln x\) is shifted vertically upwards by 3 units.
1. **Vertical Asymptote**: Begin by drawing a vertical asymptote at \(x = 0\). Remember, the graph will approach this line but never intersect it.2. **X-Intercept**: Place a point at \(x = e^{-3}\) or \(x \approx 0.0498\). This is where the graph crosses the \(x\)-axis.3. **General Shape**: The foundational shape of \(\ln x\) is an upward curve, starting near the asymptote going upwards. Because of the \(+3\) shift, simply move this curve up so that every y-value is increased by 3.

• The graph starts near \(-∞\) when \(x\) approaches zero.
• Then, it crosses the \(x\)-axis at the calculated intercept.
• Finally, it continues to rise as \(x\) increases.

Using these steps, you'll be able to sketch this logarithmic function smoothly. Remember to label all your findings clearly on the graph for accuracy!