To graph the logarithmic function [tex]\( g(x) = -\log_3 x + 1 \)[/tex], follow these steps:
### Plotting Points on the Graph
1. Choose a point to substitute into the function:
- Let's take [tex]\( x = 1 \)[/tex]:
[tex]\[
y = -\log_3(1) + 1
\][/tex]
Since [tex]\(\log_3(1) = 0\)[/tex]:
[tex]\[
y = -0 + 1 = 1
\][/tex]
Thus, the point [tex]\( (1, 1) \)[/tex] is on the graph.
2. Choose another point:
- Let's take [tex]\( x = 3 \)[/tex]:
[tex]\[
y = -\log_3(3) + 1
\][/tex]
Since [tex]\(\log_3(3) = 1\)[/tex]:
[tex]\[
y = -1 + 1 = 0
\][/tex]
Thus, the point [tex]\( (3, 0) \)[/tex] is on the graph.
### Drawing the Asymptote
Logarithmic functions have a vertical asymptote where the argument of the logarithm equals zero. For [tex]\( g(x) = -\log_3 x + 1 \)[/tex], the vertical asymptote is at [tex]\( x = 0 \)[/tex].
### Graphing the Function
1. Plot the points [tex]\( (1, 1) \)[/tex] and [tex]\( (3, 0) \)[/tex] on the coordinate plane.
2. Draw the vertical asymptote at [tex]\( x = 0 \)[/tex].
3. Sketch the curve of the function [tex]\( g(x) = -\log_3 x + 1 \)[/tex]. The curve will pass through the plotted points and will get infinitely close to the asymptote at [tex]\( x = 0 \)[/tex] without touching it.
### Domain and Range
- Domain: The domain of the function [tex]\( g(x) = -\log_3 x + 1 \)[/tex] consists of all positive real numbers [tex]\( x \)[/tex]. In interval notation, this is:
[tex]\[
(0, \infty)
\][/tex]
- Range: The range of [tex]\( g(x) = -\log_3 x + 1 \)[/tex] is all real numbers because as [tex]\( x \)[/tex] approaches [tex]\( 0^+ \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex], and as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( -\infty \)[/tex]. Hence, the range is:
[tex]\[
(-\infty, \infty)
\][/tex]
So, to fill in the details for the problem:
- Domain: [tex]\((0, \infty)\)[/tex]
- Range: [tex]\((-\infty, \infty)\)[/tex]
