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Sure! Let's analyze the function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] step by step.
### 4.1: Is [tex]\( f \)[/tex] an increasing or decreasing function? Give a reason for your answer.
The function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] is a decreasing function. This is because the base of the exponential function, [tex]\(\frac{1}{3}\)[/tex], is less than 1. In general, for exponential functions of the form [tex]\( a^x \)[/tex] where [tex]\( 0 < a < 1 \)[/tex], the function is decreasing.
### 4.2: Determine [tex]\( f^{-1}(x) \)[/tex] in the form [tex]\( y = \ldots \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex], we start by setting [tex]\( y = f^{-1}(x) \)[/tex]. This means:
[tex]\[ x = \left(\frac{1}{3}\right)^y \][/tex]
We need to solve this equation for [tex]\( y \)[/tex]. Taking the natural logarithm on both sides, we get:
[tex]\[ \ln(x) = \ln\left(\left(\frac{1}{3}\right)^y\right) \][/tex]
[tex]\[ \ln(x) = y \cdot \ln\left(\frac{1}{3}\right) \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{\ln(x)}{\ln\left(\frac{1}{3}\right)} \][/tex]
Since [tex]\( \ln\left(\frac{1}{3}\right) = -\ln(3) \)[/tex]:
[tex]\[ y = \frac{\ln(x)}{-\ln(3)} \][/tex]
Simplifying further, we get:
[tex]\[ y = -\frac{\ln(x)}{\ln(3)} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = -\frac{\ln(x)}{\ln(3)} \][/tex]
### 4.3: Write down the equation of the asymptote of [tex]\( f(x) - 5 \)[/tex].
For the function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex], the horizontal asymptote is [tex]\( y = 0 \)[/tex] because as [tex]\( x \)[/tex] approaches infinity, [tex]\( \left(\frac{1}{3}\right)^x \)[/tex] approaches 0. When we consider [tex]\( f(x) - 5 \)[/tex], this shifts the entire graph of [tex]\( f(x) \)[/tex] down by 5 units. Therefore, the equation of the asymptote for [tex]\( f(x) - 5 \)[/tex] is:
[tex]\[ y = -5 \][/tex]
### 4.4: Describe the transformation from [tex]\( f \)[/tex] to [tex]\( g \)[/tex] if [tex]\( g(x) = \log_3 x \)[/tex].
The transformation from [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] to [tex]\( g(x) = \log_3 x \)[/tex] can be described as a reflection over the y-axis. The function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] is equivalent to [tex]\( 3^{-x} \)[/tex]. Taking the logarithm base 3 of both sides, we obtain:
[tex]\[ \log_3(f(x)) = \log_3(3^{-x}) \][/tex]
[tex]\[ \log_3(f(x)) = -x \][/tex]
Thus, [tex]\( g(x) = -\log_3(f(x)) \)[/tex], making [tex]\( g(x) \)[/tex] the reflection of [tex]\( f(x) \)[/tex] over the y-axis.
In conclusion:
- 4.1 [tex]\( f \)[/tex] is a decreasing function because the base [tex]\(\frac{1}{3}\)[/tex] is less than 1.
- 4.2 [tex]\( f^{-1}(x) = -\frac{\ln(x)}{\ln(3)} \)[/tex]
- 4.3 The equation of the asymptote of [tex]\( f(x)-5 \)[/tex] is [tex]\( y = -5 \)[/tex].
- 4.4 The function [tex]\( g(x) = \log_3 x \)[/tex] is the reflection of [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] over the y-axis.
### 4.1: Is [tex]\( f \)[/tex] an increasing or decreasing function? Give a reason for your answer.
The function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] is a decreasing function. This is because the base of the exponential function, [tex]\(\frac{1}{3}\)[/tex], is less than 1. In general, for exponential functions of the form [tex]\( a^x \)[/tex] where [tex]\( 0 < a < 1 \)[/tex], the function is decreasing.
### 4.2: Determine [tex]\( f^{-1}(x) \)[/tex] in the form [tex]\( y = \ldots \)[/tex]
To find the inverse function [tex]\( f^{-1}(x) \)[/tex], we start by setting [tex]\( y = f^{-1}(x) \)[/tex]. This means:
[tex]\[ x = \left(\frac{1}{3}\right)^y \][/tex]
We need to solve this equation for [tex]\( y \)[/tex]. Taking the natural logarithm on both sides, we get:
[tex]\[ \ln(x) = \ln\left(\left(\frac{1}{3}\right)^y\right) \][/tex]
[tex]\[ \ln(x) = y \cdot \ln\left(\frac{1}{3}\right) \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{\ln(x)}{\ln\left(\frac{1}{3}\right)} \][/tex]
Since [tex]\( \ln\left(\frac{1}{3}\right) = -\ln(3) \)[/tex]:
[tex]\[ y = \frac{\ln(x)}{-\ln(3)} \][/tex]
Simplifying further, we get:
[tex]\[ y = -\frac{\ln(x)}{\ln(3)} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = -\frac{\ln(x)}{\ln(3)} \][/tex]
### 4.3: Write down the equation of the asymptote of [tex]\( f(x) - 5 \)[/tex].
For the function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex], the horizontal asymptote is [tex]\( y = 0 \)[/tex] because as [tex]\( x \)[/tex] approaches infinity, [tex]\( \left(\frac{1}{3}\right)^x \)[/tex] approaches 0. When we consider [tex]\( f(x) - 5 \)[/tex], this shifts the entire graph of [tex]\( f(x) \)[/tex] down by 5 units. Therefore, the equation of the asymptote for [tex]\( f(x) - 5 \)[/tex] is:
[tex]\[ y = -5 \][/tex]
### 4.4: Describe the transformation from [tex]\( f \)[/tex] to [tex]\( g \)[/tex] if [tex]\( g(x) = \log_3 x \)[/tex].
The transformation from [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] to [tex]\( g(x) = \log_3 x \)[/tex] can be described as a reflection over the y-axis. The function [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] is equivalent to [tex]\( 3^{-x} \)[/tex]. Taking the logarithm base 3 of both sides, we obtain:
[tex]\[ \log_3(f(x)) = \log_3(3^{-x}) \][/tex]
[tex]\[ \log_3(f(x)) = -x \][/tex]
Thus, [tex]\( g(x) = -\log_3(f(x)) \)[/tex], making [tex]\( g(x) \)[/tex] the reflection of [tex]\( f(x) \)[/tex] over the y-axis.
In conclusion:
- 4.1 [tex]\( f \)[/tex] is a decreasing function because the base [tex]\(\frac{1}{3}\)[/tex] is less than 1.
- 4.2 [tex]\( f^{-1}(x) = -\frac{\ln(x)}{\ln(3)} \)[/tex]
- 4.3 The equation of the asymptote of [tex]\( f(x)-5 \)[/tex] is [tex]\( y = -5 \)[/tex].
- 4.4 The function [tex]\( g(x) = \log_3 x \)[/tex] is the reflection of [tex]\( f(x) = \left(\frac{1}{3}\right)^x \)[/tex] over the y-axis.
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