I believe that the domain for f(x) = log subscript 3 (1-x) is x is a real number I think the range for the same equation is y > 1 .
the domain of log (1-x) would be such that
1-x > 0
-x>-1
x < 1
whereas the range would be y is any real number
Have you considered what the graph would look like?
Yes, I have considered what the graph of the function f(x) = log₃(1-x) would look like. To better understand the domain and range of this function, let's break it down step by step.
First, let's consider the domain, which represents all possible values that x can take. In this case, we have the function f(x) = log₃(1-x).
To determine the domain, we need to look at the argument of the logarithm function, which is (1-x). The domain of a logarithm function excludes any values that would make the argument negative or equal to zero.
In this case, we have (1-x) as the argument. Therefore, for (1-x) to be positive, we need:
1 - x > 0
We can solve this inequality as follows:
x < 1
Based on this, the domain of the function is all x values such that x is less than 1.
Moving on to the range, which represents all possible values that y (or f(x)) can take. A logarithm function can take any real number input and transform it into a real number output. Therefore, the range of the logarithm function f(x) = log₃(1-x) is all real numbers.
In conclusion, the domain of the function f(x) = log₃(1-x) is x < 1 and the range is all real numbers.
As for the graph, the logarithmic function f(x) = log₃(1-x) will have a vertical asymptote at x = 1, meaning that the function will approach infinity as x approaches 1 from the left side. The graph will also be decreasing and will approach negative infinity as x approaches negative infinity.