Problem 68 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The domain of the given function is \(x < 3\). The function crosses the x-axis at \(x = 2\), and there's a vertical asymptote at \(x = 3\).

Step 1: Find the Domain of the Function

The domain of a logarithmic function \(f(x) = ln(x)\) is for all values \(x > 0\). Therefore, for the function \(f(x)=\ln (3-x)\), the values of x must satisfy \(3 - x > 0\). Solving this inequality for x gives \(x < 3\). So the domain of the function is \(x < 3\).

Step 2: Find the X-intercepts

The function \(f(x)\) will cross the x-axis wherever \(f(x) = 0\). So, setting \(f(x) = 0\) for our function gives \(0 = ln(3-x)\) Exploitation of the property of the logarithm that \(e^{ln(x)} = x\), we find that \( e^{0} = 3 - x \). Since \(e^{0}= 1\), we solve for \(x\) to give \( x = 3 - 1 = 2\). Thus, the function crosses the x-axis at \(x = 2\).

Step 3: Find the Vertical Asymptotes

Vertical asymptotes of logarithmic functions occur wherever the argument of the logarithm function becomes zero. For our function, this means setting \(3 - x = 0\) and solving for \(x\), which results in \(x = 3\). So the function has a vertical asymptote at \(x = 3\).

Step 4: Sketch of the Graph

To draw the graph, plot the x-intercept at \(x = 2\) and plot the vertical asymptote as a dotted line at x = 3. Then sketch the graph of the function knowing that it is a logarithmic function that approaches the asymptote as X goes towards 3, but never quite reaches it. The function will cross the x-axis at \(x = 2\) and will continue to increase slowly as \(x\) moves away from 3 but less than 3.

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial, as it tells us all the possible values that make the function work. For a logarithmic function like \( f(x) = \ln(3-x) \), the argument inside the log, \(3-x\), must be greater than zero. This is because logarithms of non-positive numbers are undefined in the real number system.
\[3-x > 0\]
Solving for \(x\) gives:
\[x < 3\]
Therefore, the domain of this function is all real numbers less than 3. This means you can plug in any value for \(x\) that is smaller than 3, and the function will give you a real output. To visualize, imagine a number line where every point to the left of 3 is included, up to but not including 3.

Vertical Asymptotes

Vertical asymptotes in logarithmic functions occur where the argument of the log equals zero. This represents a point where the function shoots off to infinity and is an important feature in graphing. For \( f(x) = \ln(3-x) \), set the argument equal to zero:
\[3-x = 0\] Solving this gives \(x = 3\).
Thus, the vertical asymptote is at \(x = 3\).

Here's how it looks on a graph:

• The function will approach this line as \(x\) gets closer to 3 from the left side.
• However, it will never actually touch or cross the line \(x = 3\).

Imagine a barrier at \(x = 3\), where the function tries to reach but can't, creating an infinite climb.

X-intercepts

To find where the function crosses the x-axis, look for the x-intercept. This is found by setting \(f(x) = 0\):
\[0 = \ln(3-x)\] Using the property \(e^{\ln(x)} = x\), we convert:
\[ e^0 = 3 - x \] As \(e^0 = 1\), solve for \(x\):
\[ 3 - x = 1 \]
\[ x = 2 \]
So, the x-intercept is at \(x = 2\).

Why is this important? Because it's the point where the function actually touches the x-axis. Some key points about x-intercepts:

• The function equals zero here.
• For \(f(x) = \ln(3-x)\), it means the graph crosses the x-axis at \(x = 2\).

This gives us a clear visual anchor point when sketching the graph.