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onsider the function \( f(x)=\left(\frac{1}{3}\right)^{x} \). . Is \( f \) an increasing or decreasing function? Give a reason for your anower .2 Determine \( f^{-1}(x) \) in the form \( y=- \) .3 Write down the equation of the asymptote of \( f(x)-5 \). .4 Describe the transformation from \( f \) in \( g \) if \( g(x)=\log _{3} x \).

Ask by Schofield O'Quinn. in South Africa
Feb 25,2025

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Answer

1. The function \( f(x) = \left(\frac{1}{3}\right)^{x} \) is **decreasing** because the base \( \frac{1}{3} \) is less than 1. 2. The inverse function is \( f^{-1}(x) = -\log_{3}(x) \). 3. The horizontal asymptote of \( f(x) - 5 \) is \( y = -5 \). 4. The transformation from \( f(x) \) to \( g(x) = \log_{3}(x) \) is a **reflection across the line \( y = x \)**.

Solution

Let's solve the problem step by step. ### 1. Determine if \( f(x) \) is increasing or decreasing The function given is \( f(x) = \left(\frac{1}{3}\right)^{x} \). To determine if the function is increasing or decreasing, we can analyze the base of the exponential function. Since \( \frac{1}{3} < 1 \), the function is a decreasing exponential function. **Reason**: In general, if \( 0 < a < 1 \), then \( f(x) = a^x \) is a decreasing function because as \( x \) increases, \( a^x \) decreases. ### 2. Determine \( f^{-1}(x) \) To find the inverse function \( f^{-1}(x) \), we start with the equation: \[ y = \left(\frac{1}{3}\right)^{x} \] Now, we will solve for \( x \) in terms of \( y \): 1. Take the logarithm of both sides: \[ \log_{3}(y) = \log_{3}\left(\left(\frac{1}{3}\right)^{x}\right) \] 2. Using the property of logarithms, we can simplify: \[ \log_{3}(y) = x \cdot \log_{3}\left(\frac{1}{3}\right) \] 3. Since \( \log_{3}\left(\frac{1}{3}\right) = -1 \): \[ \log_{3}(y) = -x \] 4. Rearranging gives: \[ x = -\log_{3}(y) \] Thus, the inverse function is: \[ f^{-1}(x) = -\log_{3}(x) \] ### 3. Write down the equation of the asymptote of \( f(x) - 5 \) The function \( f(x) = \left(\frac{1}{3}\right)^{x} \) approaches \( 0 \) as \( x \) approaches infinity. Therefore, the horizontal asymptote of \( f(x) \) is \( y = 0 \). For the function \( f(x) - 5 \): \[ f(x) - 5 = \left(\frac{1}{3}\right)^{x} - 5 \] The horizontal asymptote will shift down by 5 units. Thus, the equation of the asymptote is: \[ y = -5 \] ### 4. Describe the transformation from \( f \) to \( g \) where \( g(x) = \log_{3}(x) \) The function \( f(x) = \left(\frac{1}{3}\right)^{x} \) is an exponential function, while \( g(x) = \log_{3}(x) \) is a logarithmic function. The transformation from \( f \) to \( g \) can be described as follows: - The function \( g(x) \) is the inverse of \( f(x) \). - Specifically, \( g(x) \) represents a reflection of \( f(x) \) across the line \( y = x \). In summary, the transformation from \( f \) to \( g \) is a reflection across the line \( y = x \). ### Final Summary 1. \( f(x) \) is a decreasing function. 2. \( f^{-1}(x) = -\log_{3}(x) \). 3. The asymptote of \( f(x) - 5 \) is \( y = -5 \). 4. The transformation from \( f \) to \( g \) is a reflection across the line \( y = x \).

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Bonus Knowledge

The function \( f(x) = \left(\frac{1}{3}\right)^{x} \) is a decreasing function because as \( x \) increases, the base \( \frac{1}{3} \) raised to a higher power results in smaller values. Essentially, we're flipping the numbers: smaller bases with exponents yield lesser outputs, making the function slide downwards as you move along the x-axis. To find the inverse \( f^{-1}(x) \), start by setting \( y = f(x) = \left(\frac{1}{3}\right)^{x} \), then solve for \( x \): we can rewrite it as \( \log_{\frac{1}{3}}(y) = x \). Changing the base gives us \( x = -\log_{3}(y) \), hence \( f^{-1}(x) = -\log_{3}(x) \). The equation of the asymptote of \( f(x) - 5 \) is \( y = -5 \). This comes from observing that \( f(x) \) approaches 0 as \( x \) heads towards infinity, meaning subtracting 5 shifts this horizontal line downwards without affecting the direction of approach. The transformation from \( f(x) \) to \( g(x) = \log_{3} x \) can be understood as a reflection and a shift. The logarithmic function represents the inverse of exponential decay, indicating that as the output of \( f(x) \) decreases, the input \( g(x) \) effectively captures that behavior by shifting our perspective from y-values into x-values, making it a fascinating flip from one function type to another!

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