Problem 41 - A Graphical Approach to College Algebra

Question

Graph each function. $$f(x)=\log _{1 / 2}(1-x)$$

Step-by-Step Solution

Answer

The graph of \( f(x) = \log_{1/2}(1-x) \) decreases from the left, crossing the x-axis at \( x=0 \), with a vertical asymptote at \( x=1 \).

Step 1: Understand the domain of the function

For the logarithmic function \( f(x) = \log_{1/2}(1-x) \), the argument \( 1-x \) must be positive. This means we need \( 1-x > 0 \), which simplifies to \( x < 1 \). Therefore, the domain of \( f(x) \) is all real numbers \( x \) such that \( x < 1 \).

Step 2: Identify asymptotes and intercepts

We know the function is undefined at \( x = 1 \), which makes \( x = 1 \) a vertical asymptote. The function crosses the x-axis when \( \log_{1/2}(1-x) = 0 \), meaning \( 1-x = 1 \). This gives \( x = 0 \), which provides the x-intercept. Since \( f(x) \) doesn't include a \( y \)-intercept as it’s only defined for \( x < 1 \), no \( y \)-intercept we consider.

Step 3: Analyze the behavior near the asymptote

As \( x \) approaches 1 from the left (\( x \to 1^- \)), \( 1-x \to 0^+ \), and since the base \( 1/2 < 1 \), the logarithm goes to negative infinity: \( \log_{1/2}(1-x) \to -\infty \).

Step 4: Choose key points to plot

Choose a few points in the domain to evaluate and plot: \( f(0) = \log_{1/2}(1) = 0 \), \( f(-1) = \log_{1/2}(2) = -1 \), \( f(-3) = \log_{1/2}(4) = -2 \). Calculate and plot these points on the graph.

Step 5: Sketch the graph

Plot the key points on a graph, keeping in mind that \( f(x) \) approaches \( -\infty \) as \( x \to 1^- \). Connect these points smoothly, indicating the vertical asymptote at \( x = 1 \). Ensure the graph decreases as you move from left to right up to the asymptote.

Key Concepts

Domain and Range

A logarithmic function like \( f(x) = \log_{1/2}(1-x) \) has specific requirements for its domain and range. The argument of the logarithm, in this case \( 1-x \), must be positive. Hence, for \( f(x) \) to be defined, \( 1-x > 0 \), simplifying to \( x < 1 \). Thus, the domain for this function is all real numbers where \( x < 1 \).
The range of logarithmic functions is generally all real numbers. This is because as \( x \) approaches the lower part of its domain (in this case near 1, just under it), the function value can decrease toward \(-\infty\). As \( x \) decreases further to very negative values, the function value can also increase to very high numbers. Therefore, we can say the range of \( f(x) \) is all real numbers from \(-\infty\) to \( \infty \).
Understanding these constraints is crucial as they affect how we approach graphing the function effectively.

Asymptotes

Logarithmic functions often have vertical asymptotes, helping us understand where the graph shoots off towards infinity. In this function, \( f(x) = \log_{1/2}(1-x) \), there is a vertical asymptote where the function becomes undefined. This means looking at the point where \( 1-x = 0 \), which gives \( x = 1 \).
As \( x \) approaches 1 from the left (denoted as \( x \to 1^- \)), the value of \( 1-x \) approaches zero. Since the base of the logarithm \( 1/2 \) is less than 1, \( \log_{1/2}(y) \) diverges toward \(-\infty\) as \( y \to 0^+ \). Hence, the line \( x = 1 \) is a vertical asymptote and the function decreases toward \(-\infty\) as \( x \) approaches this value. Vertical asymptotes are crucial markers when graphing as they determine where the curve extends off the graph without ever touching a certain line.

Graphing Techniques

Graphing \( f(x) = \log_{1/2}(1-x) \) involves using strategic points and understanding the behavior of logarithmic functions. First, determining key x-intercepts like where \( \log_{1/2}(1-x) = 0 \) gives us \( 1-x = 1 \), leading to \( x = 0 \). This x-intercept implies that the curve crosses the x-axis at \( x = 0 \).
To plot the graph accurately, evaluate \( f(x) \) at various points within the domain such as \( f(0) = 0 \), \( f(-1) = \log_{1/2}(2) = -1 \), and \( f(-3) = \log_{1/2}(4) = -2 \). Plotting these points reveals the downward slope as \( x \) goes from left to right, approaching the asymptote. Remember that with \( x \to 1^- \), \( f(x) \rightarrow -\infty \), pulling the graph steeply downward near the asymptote.
Connect these points with a smooth curve keeping the asymptote in mind, reflecting a decrease as x increases toward the vertical asymptote. This provides a holistic graphical representation of the entire function, essential for visualizing how logarithmic functions behave in real contexts.