Problem 52 - A Graphical Approach to Precalculus with Limits

Question

For each exponential function \(f\), find \(f^{-1}\) analytically and graph \(f\) and \(f^{-1}\) as \(Y_{1}\) and \(Y_{2}\) in the same viewing window. $$f(x)=\left(\frac{1}{2}\right)^{x}-5$$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = -\log_{2}(x + 5) \).

Step 1: Set the Function Equal to y

Begin by letting the function be equal to \( y \). Therefore, we have:\[y = \left(\frac{1}{2}\right)^{x} - 5\]

Step 2: Solve for x

To find \( f^{-1} \), solve the equation from Step 1 for \( x \). First, add 5 to both sides:\[y + 5 = \left(\frac{1}{2}\right)^{x}\]

Step 3: Convert Exponential to Logarithmic Form

To solve for \( x \), convert the exponential equation to logarithmic form. Recall that if \( a^b = c \), then \( b = \log_a(c) \). Thus, \[\x = \log_{\frac{1}{2}}(y + 5)\]

Step 4: Simplify the Logarithm

Since we are dealing with base \( \frac{1}{2} \), the logarithm can be rewritten using base 2:\[\x = \frac{\log_{2}(y + 5)}{-1}\]Thus, \[\x = -\log_{2}(y + 5)\]

Step 5: Swap x and y to Find f^{-1}

To find the inverse function, swap \( x \) and \( y \):\[\y = -\log_{2}(x + 5)\]Therefore, the inverse function is \[\ f^{-1}(x) = -\log_{2}(x + 5)\]

Step 6: Discuss Graphing f(x) and f^{-1}(x)

To graph \( f(x) = \left(\frac{1}{2}\right)^{x} - 5 \) and its inverse \( f^{-1}(x) = -\log_{2}(x + 5) \), plot \( f(x) \) and \( f^{-1}(x) \) on the same axis. Remember that graphing an exponential function typically involves plotting points and recognizing it decreases rapidly for \( x \to +\infty \). Meanwhile, the inverse should be symmetric to the original function across the line \( y = x \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They have the general form of \( f(x) = a^{x} \), where \( a \) is a positive constant. This class of functions is significant because their rate of growth (or decay) is proportional to their value. This leads them to either increase or decrease rapidly. In our given problem, the exponential function is \( f(x) = \left(\frac{1}{2}\right)^{x} - 5 \). Here, the base \( \frac{1}{2} \) indicates this function involves exponential decay as \( x \) increases.

Some key characteristics of exponential functions include:

• Domain: All real numbers (\(-\infty, \infty\)).
• Range: Depends on the function's form, but for \( \left(\frac{1}{2}\right)^x - 5 \), it is \((-5, \infty)\).
• Intercept: Often crosses the y-axis at a point easily derived by evaluating \( f(0) \).
• Shape: Either increases or decreases rapidly, depending on whether the base is greater than or less than 1.

Explaining these functions helps in understanding their inverses, which involve solving equations that shift resources from the exponential to the logarithmic form.

Logarithmic Functions

Logarithmic functions, often viewed as the inverses of exponential functions, assist in solving equations where the variable is an exponent. For an exponential equation like \( a^x = b \), expressing \( x \) in terms of \( b \) leads to \( x = \log_a(b) \). For our problem, the inverse of the exponential function \( \left(\frac{1}{2}\right)^{x} - 5 \) involves solving for \( x \) by setting \( y + 5 = \left(\frac{1}{2}\right)^{x} \) and rewriting it as a logarithmic equation: \( x = -\log_2(y + 5) \).Some important points for understanding logarithmic functions include:

• Inverse Nature: Logarithms undo the operation of exponentiation.
• Base Correspondence: The base of the logarithm corresponds to the base of the exponential function, thus \( \log_{\frac{1}{2}} \) is involved here.
• Domain and Range: These functions often have a domain of all positive numbers and a range of all real numbers, but they change based on transformations applied.

In the context of our inverse function \( f^{-1}(x) = -\log_{2}(x + 5) \), understanding that the negative sign reflects the graph is crucial in graphing and interpreting it.

Function Graphing

Graphing functions, and particularly their inverses, can help visual learners grasp mathematical scenarios better. For our scenario, we have two main functions to graph: the exponential function \( f(x) = \left(\frac{1}{2}\right)^{x} - 5 \) and its inverse \( f^{-1}(x) = -\log_2(x + 5) \). Some fundamental tips for graphing exponential and logarithmic functions include:

• Symmetry: The function and its inverse are symmetric about the line \( y = x \). This line acts as a mirror creating a visually identifiable pattern between the graphs.
• Intersections: Plotting the y-intercept and specific points carefully will guide your sketch. Notably, the point of intersection of \( f(x) \) with the x-axis or y-axis provides important insight for graphing.
• Behavior at Extremes: Exponential functions typically rise or fall swiftly, moving towards asymptotes. Conversely, logarithmic graphs rise slowly and start from a defined point.

When both function graphs are placed on the same axes, the student's ability to visualize the inverse relationship simply strengthens. Recognizing how these functions interact provides a practical understanding of their broader applications.