Question

Use the given function \(f\) (a) Find the domain of \(f\). (b) Graph \(f\) (c) From the graph, determine the range and any asymptotes of \(f\) (d) Fund \(f^{-1}\), the tnverse of \(f\). (e) Find the domain and the range of \(f^{-1}\). (f) Graph \(f^{-1}\). $$ f(x)=2-\log _{3}(x+1) $$

Short answer

Domain of \( f(x) \): \((-1, \, \infty)\). Range: \((-\infty, \, 2)\). Vertical asymptote: \(x = -1\). Inverse function: \(f^{-1}(x) = 3^{2 - x} - 1\). Domain of \(f^{-1}(x)\): \((-\infty, \, 2)\). Range of \(f^{-1}(x)\): \((-1, \, \infty)\).

Step-by-step solution

- Find the Domain of f(x)

The domain of a logarithmic function is found by setting the argument of the logarithm greater than zero. For the given function:\[ x + 1 > 0 \]Solving this inequality gives:\[ x > -1 \]So, the domain of \( f(x) \) is \( (-1, \, \infty) \).

- Graph f(x)

To graph \( f(x) = 2 - \log_{3}(x+1) \), plot some key points. For example:x = 0: \( f(0) = 2 - \log_{3}(1) = 2 \)x = 2: \( f(2) = 2 - \log_{3}(3) = 1 \)x = 8: \( f(8) = 2 - \log_{3}(9) = 2 - 2 = 0 \)Notice that there's a vertical asymptote at \( x = -1 \), as the logarithm is undefined at that point.

- Determine the Range and Asymptotes of f(x)

From the graph, it is clear that as x approaches \( -1 \) from the right, \( f(x) \) goes to infinity. As x approaches infinity, \( f(x) \) decreases without bound. Therefore, the range is \( (-\infty, \, 2) \). Additionally, there is a vertical asymptote at \( x = -1 \).

- Find the Inverse Function

To find the inverse of \( f(x) = 2 - \log_{3}(x+1) \):1. Replace \( f(x) \) with y: \[ y = 2 - \log_{3}(x+1) \]2. Solve for x in terms of y: \[ y - 2 = -\log_{3}(x+1) \] \[ -(y - 2) = \log_{3}(x+1) \] \[ x + 1 = 3^{- (y - 2)} \] \[ x = 3^{2 - y} - 1 \]3. Swap x and y to get the inverse function: \[ f^{-1}(x) = 3^{2 - x} - 1 \].

- Find the Domain and Range of f^{-1}(x)

The domain of \( f^{-1}(x) \) is the range of \( f(x) \), which is \( (-\infty, \, 2) \).The range of \( f^{-1}(x) \) is the domain of \( f(x) \), which is \( (-1, \, \infty) \).

- Graph f^{-1}(x)

To graph \( f^{-1}(x) = 3^{2 - x} - 1 \), plot key points similar to the process in Step 2. For example:x = 2: \( f^{-1}(2) = 3^{0} - 1 = 0 \)x = 0: \( f^{-1}(0) = 3^{2} - 1 = 8 \)x = -1: \( f^{-1}(-1) = 3^{3} - 1 = 26 \)Ensure to include the appropriate vertical and horizontal asymptotes. The horizontal asymptote here is when \( y= -1 \).

Key concepts

Domain and Range

The domain of a function is the set of all possible input values (x-values) that the function can accept,
while the range is the set of all possible output values (y-values) that the function can produce.

For the given function, \( f(x) = 2 - \log\textsubscript{3}(x+1)\), we find the domain by setting the argument of the logarithm greater than zero:
\[ x + 1 > 0 \] Solving the inequality, we get:
\[ x > -1 \] So, the domain of \( f(x) \) is \( (-1, \infty) \).

The range of the function can be determined from its graph. As \( x \) approaches \( -1 \) from the right, \( f(x) \) goes to infinity, and as \( x \) goes to infinity, \( f(x) \) decreases without bound.
Therefore, the range of \( f(x) \) is \( (-\infty, 2) \).

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions.
For a logarithmic function of the form \( \log\textsubscript{b}(y) = x \), where \( b \) is the base, it can be rewritten in its exponential form as \( y = b\textsuperscript{x} \).

In the given function \( f(x) = 2 - \log\textsubscript{3}(x+1) \), it's important to note how the logarithm affects the values as \( x \) changes.
The subtraction of the logarithmic term \( \log\textsubscript{3}(x+1) \) from 2 shifts the graph of the base-3 logarithm downward.
Additionally, logarithmic functions have vertical asymptotes where their arguments approach zero,
which will influence both the graph and the range of the function.
For instance, as \( x \) approaches -1, the logarithmic term becomes undefined, causing a vertical asymptote at \( x = -1 \).

Graphing Transformations

Graphing transformations involve shifting, reflecting, stretching, or compressing the graph of a given function.
In \( f(x) = 2 - \log\textsubscript{3}(x+1) \), several transformations occur:

• The \( \log\textsubscript{3}(x+1) \) function is shifted left by 1 unit due to \( (x + 1) \).
  This means the typical vertical asymptote \( x = 0 \) of \( \log\textsubscript{3}(x) \) moves to \( x = -1 \).
• The negative sign before the \( \log \) reflects the graph across the horizontal axis.
• Adding 2 shifts the entire graph up by 2 units

To graph \( f(x) \):

• Start with the base graph of \( \log\textsubscript{3}(x+1) \).
• Apply the vertical reflection across the x-axis.
• Shift the graph up 2 units.
• Note the vertical asymptote at \( x = -1 \).

Asymptotes

Asymptotes are lines that the graph of a function approaches but never touches.
They can be vertical, horizontal, or oblique. They provide key information about the behavior of the function at the extremes.

In the function \( f(x) = 2 - \log\textsubscript{3}(x+1) \), there is a vertical asymptote at \( x = -1 \),
because the logarithmic function is undefined when the argument \( x+1= 0 \).
This makes the output of the function go to infinity as \( x \) approaches -1 from the right.
The graph also indicates the horizontal behavior, showing that as \( x \) approaches infinity, \( f(x) \) decreases without bound.
Therefore, there is no horizontal asymptote for this function,
but there is significant behavior change as \( x \) extends (the output values approach \( - \infty \)).
When graphing the inverse function \( f\textsuperscript{-1}(x) = 3\textsuperscript{2-x} - 1 \), it has a horizontal asymptote at \( y = -1 \).