Answer :
Let's consider the function [tex]\( g: x \mapsto \log_3(x + 2) - 2 \)[/tex].
### a. Transformation
To transform the function [tex]\( y = \log_3(x) \)[/tex] to [tex]\( y = \log_3(x + 2) - 2 \)[/tex], we perform the following steps:
1. Horizontal shift: [tex]\( \log_3(x) \)[/tex] to [tex]\( \log_3(x + 2) \)[/tex]. This represents a shift to the left by 2 units.
2. Vertical shift: [tex]\( \log_3(x + 2) \)[/tex] to [tex]\( \log_3(x + 2) - 2 \)[/tex]. This represents a shift down by 2 units.
Thus, the transformation is: Horizontal shift of 2 units to the left and vertical shift of 2 units downward.
### b. Domain and Range
Domain:
The argument of the logarithmic function, [tex]\( x + 2 \)[/tex], must be positive:
[tex]\[ x + 2 > 0 \][/tex]
[tex]\[ x > -2 \][/tex]
Therefore, the domain of [tex]\( g(x) \)[/tex] is [tex]\( (-2, \infty) \)[/tex].
Range:
Since the range of [tex]\( \log_3(x) \)[/tex] is all real numbers and we are only shifting vertically by a constant, the range of [tex]\( g(x) \)[/tex] remains all real numbers:
[tex]\[ (-\infty, \infty) \][/tex]
### c. Asymptotes and Intercepts
Asymptotes:
The vertical asymptote occurs where the argument of the logarithm goes to zero:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -2 \)[/tex].
Intercepts:
- Y-intercept: Set [tex]\( x = 0 \)[/tex] in [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = \log_3(0 + 2) - 2 = \log_3(2) - 2 \][/tex]
- X-intercept: Solve [tex]\( g(x) = 0 \)[/tex]:
[tex]\[ \log_3(x + 2) - 2 = 0 \][/tex]
[tex]\[ \log_3(x + 2) = 2 \][/tex]
[tex]\[ x + 2 = 3^2 = 9 \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, the intercepts are:
- [tex]\( y \)[/tex]-intercept: [tex]\( \left(0, \log_3(2) - 2 \right) \)[/tex]
- [tex]\( x \)[/tex]-intercept: [tex]\( (7, 0) \)[/tex]
### d. Inverse Function
To find the inverse function [tex]\( g^{-1} \)[/tex]:
1. Start with [tex]\( y = \log_3(x + 2) - 2 \)[/tex].
2. Swap [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ x = \log_3(y + 2) - 2 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x + 2 = \log_3(y + 2) \][/tex]
[tex]\[ 3^{x + 2} = y + 2 \][/tex]
[tex]\[ y = 3^{x + 2} - 2 \][/tex]
Thus, the inverse function is:
[tex]\[ g^{-1}(x) = 3^{x + 2} - 2 \][/tex]
### e. Sketch the Graphs
To sketch the graphs of [tex]\( g(x) \)[/tex], [tex]\( g^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex]:
1. Graph of [tex]\( g(x) = \log_3(x + 2) - 2 \)[/tex]:
- Vertical asymptote at [tex]\( x = -2 \)[/tex].
- Passes through intercepts ([tex]\(0, \log_3(2) - 2\)[/tex]) and (7,0).
- General shape of a logarithmic function shifted left and down.
2. Graph of [tex]\( g^{-1}(x) = 3^{x + 2} - 2 \)[/tex]:
- This is an exponential function.
- Horizontal asymptote at [tex]\( y = -2 \)[/tex].
- Passes through (0,7) and intercepts [tex]\((\log_3(2)-2,0)\)[/tex].
3. Graph of [tex]\( y = x \)[/tex]:
- A straight line passing through the origin with a slope of 1.
When sketching, ensure the graphs intersect at points which are mutual inverses considering the functions [tex]\( g \)[/tex] and [tex]\( g^{-1} \)[/tex]. Place the following points on the graph:
- Points of intersection like (7,0) on [tex]\( g(x) \)[/tex] intersects at (0,7) on [tex]\( g^{-1}(x) \)[/tex].
- The y-intercept [tex]\(\left(0, \log_3(2) - 2 \right)\)[/tex] on [tex]\( g(x) \)[/tex] and it's corresponding inverse point.
With this information, the sketches should manifest the linked transformations and the relationship between the logarithmic and exponential functions.
### a. Transformation
To transform the function [tex]\( y = \log_3(x) \)[/tex] to [tex]\( y = \log_3(x + 2) - 2 \)[/tex], we perform the following steps:
1. Horizontal shift: [tex]\( \log_3(x) \)[/tex] to [tex]\( \log_3(x + 2) \)[/tex]. This represents a shift to the left by 2 units.
2. Vertical shift: [tex]\( \log_3(x + 2) \)[/tex] to [tex]\( \log_3(x + 2) - 2 \)[/tex]. This represents a shift down by 2 units.
Thus, the transformation is: Horizontal shift of 2 units to the left and vertical shift of 2 units downward.
### b. Domain and Range
Domain:
The argument of the logarithmic function, [tex]\( x + 2 \)[/tex], must be positive:
[tex]\[ x + 2 > 0 \][/tex]
[tex]\[ x > -2 \][/tex]
Therefore, the domain of [tex]\( g(x) \)[/tex] is [tex]\( (-2, \infty) \)[/tex].
Range:
Since the range of [tex]\( \log_3(x) \)[/tex] is all real numbers and we are only shifting vertically by a constant, the range of [tex]\( g(x) \)[/tex] remains all real numbers:
[tex]\[ (-\infty, \infty) \][/tex]
### c. Asymptotes and Intercepts
Asymptotes:
The vertical asymptote occurs where the argument of the logarithm goes to zero:
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
So, there is a vertical asymptote at [tex]\( x = -2 \)[/tex].
Intercepts:
- Y-intercept: Set [tex]\( x = 0 \)[/tex] in [tex]\( g(x) \)[/tex]:
[tex]\[ g(0) = \log_3(0 + 2) - 2 = \log_3(2) - 2 \][/tex]
- X-intercept: Solve [tex]\( g(x) = 0 \)[/tex]:
[tex]\[ \log_3(x + 2) - 2 = 0 \][/tex]
[tex]\[ \log_3(x + 2) = 2 \][/tex]
[tex]\[ x + 2 = 3^2 = 9 \][/tex]
[tex]\[ x = 7 \][/tex]
Therefore, the intercepts are:
- [tex]\( y \)[/tex]-intercept: [tex]\( \left(0, \log_3(2) - 2 \right) \)[/tex]
- [tex]\( x \)[/tex]-intercept: [tex]\( (7, 0) \)[/tex]
### d. Inverse Function
To find the inverse function [tex]\( g^{-1} \)[/tex]:
1. Start with [tex]\( y = \log_3(x + 2) - 2 \)[/tex].
2. Swap [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ x = \log_3(y + 2) - 2 \][/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x + 2 = \log_3(y + 2) \][/tex]
[tex]\[ 3^{x + 2} = y + 2 \][/tex]
[tex]\[ y = 3^{x + 2} - 2 \][/tex]
Thus, the inverse function is:
[tex]\[ g^{-1}(x) = 3^{x + 2} - 2 \][/tex]
### e. Sketch the Graphs
To sketch the graphs of [tex]\( g(x) \)[/tex], [tex]\( g^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex]:
1. Graph of [tex]\( g(x) = \log_3(x + 2) - 2 \)[/tex]:
- Vertical asymptote at [tex]\( x = -2 \)[/tex].
- Passes through intercepts ([tex]\(0, \log_3(2) - 2\)[/tex]) and (7,0).
- General shape of a logarithmic function shifted left and down.
2. Graph of [tex]\( g^{-1}(x) = 3^{x + 2} - 2 \)[/tex]:
- This is an exponential function.
- Horizontal asymptote at [tex]\( y = -2 \)[/tex].
- Passes through (0,7) and intercepts [tex]\((\log_3(2)-2,0)\)[/tex].
3. Graph of [tex]\( y = x \)[/tex]:
- A straight line passing through the origin with a slope of 1.
When sketching, ensure the graphs intersect at points which are mutual inverses considering the functions [tex]\( g \)[/tex] and [tex]\( g^{-1} \)[/tex]. Place the following points on the graph:
- Points of intersection like (7,0) on [tex]\( g(x) \)[/tex] intersects at (0,7) on [tex]\( g^{-1}(x) \)[/tex].
- The y-intercept [tex]\(\left(0, \log_3(2) - 2 \right)\)[/tex] on [tex]\( g(x) \)[/tex] and it's corresponding inverse point.
With this information, the sketches should manifest the linked transformations and the relationship between the logarithmic and exponential functions.
