Problem 112 - College Algebra with Modeling and Visualization

Question

Complete the following. (a) Graph \(y=f(x), y=f^{-1}(x),\) and \(y=x\) (b) Determine the intervals where \(f\) and \(f^{-1}\) are increasing or decreasing. $$f(x)=\log _{1 / 2} x$$

Step-by-Step Solution

Answer

Both functions are decreasing for their respective domains.

Step 1: Understand the Function

The function given is \( f(x) = \log_{1/2}(x) \). This is a logarithmic function with base 1/2. Since the base is between 0 and 1, the function is decreasing.

Step 2: Analyze the Inverse Function

The inverse of a logarithmic function \( y = \log_a(x) \) is an exponential function \( x = a^y \). Therefore, the inverse function is \( f^{-1}(x) = (1/2)^x \). This function is an exponential decay, since the base is less than 1, and it will be decreasing.

Step 3: Graph the Functions

To graph \( y = f(x) \), \( y = f^{-1}(x) \), and \( y = x \):- The graph of \( y = \log_{1/2}(x) \) will decrease from infinity to negative infinity as it goes from left to right.- The graph of \( y = (1/2)^x \) will decrease as x increases.- The line \( y = x \) is a diagonal line from bottom-left to top-right.- The graphs of \( y = f(x) \) and \( y = f^{-1}(x) \) should be symmetric about the line \( y = x \).

Step 4: Determine Intervals of Increase/Decrease for f

The function \( f(x) = \log_{1/2}(x) \) is decreasing for all \( x > 0 \). This is because logarithmic functions with bases between 0 and 1 are always decreasing.

Step 5: Determine Intervals of Increase/Decrease for f^{-1}

The function \( f^{-1}(x) = (1/2)^x \) is decreasing for all \( x \). This is characteristic of exponential functions with bases less than 1.

Key Concepts

Inverse Functions

Inverse functions are essential for understanding the relationship between inputs and outputs in a given function. To find the inverse of a function, we essentially swap the roles of the input and output. For example, the original function given is a logarithmic function:
\[ f(x) = \log_{1/2}(x) \]
To find the inverse, we consider that the inverse function should undo this operation. A logarithmic function's inverse is an exponential function. For our function, the inverse is:
\[ f^{-1}(x) = (1/2)^x \]
This illustrates a general principle: the inverse of \( y = \log_a(x) \) is \( x = a^y \). When the base is between 0 and 1, as in this exercise, both the function and its inverse are decreasing. Recognizing these pairs of operations helps in solving equations where we know either the input or output.

Graphing Functions

Graphing functions visually represents how a function behaves over a set of inputs. Graphs offer a quick way to see properties like increases, decreases, and symmetry.
In this exercise, you are asked to graph three functions:

• The original function \( y = \log_{1/2}(x) \)
• The inverse function \( y = (1/2)^x \)
• The line \( y = x \)

The key to graphing \( y = \log_{1/2}(x) \) is understanding it decreases from right to left, starting from positive infinity. Conversely, \( y = (1/2)^x \) decreases as \( x \) increases and starts near 1 when \( x \) is zero, falling towards zero. The line \( y = x \) is a mirror line showing symmetry, as the graph of a function and its inverse are symmetric about it.

Exponential Functions

Exponential functions are powerful mathematical tools used to model various real-world phenomena, like population decay or radioactive decay. They are written in the general form \( y = a^x \), where \( a \) is a constant. In this task, the inverse of the original logarithmic function is the exponential function:
\[ f^{-1}(x) = (1/2)^x \]
This is a decay function because its base, 1/2, is less than 1. As \( x \) increases, the function value decreases exponentially, approaching zero but never reaching it. This behavior is crucial for understanding decay processes, highlighting how quantities decrease over time continuously at a consistent rate. Understanding both the graph and the mechanics of this function helps in both theoretical calculations and practical interpretations.