Let f(x)=ln(x). Find the domain of the function g(x)=f(e−f(x))
e^-f(x) = e^-lnx = 1/x
so, g(x) = ln(1/x) = -lnx = -f(x)
(0, e^e)? or is it just (0,e)?
To find the domain of the function g(x) = f(e^(-f(x))), we need to consider the valid values of x that do not result in any undefined or imaginary numbers.
First, let's understand the function f(x) = ln(x). The natural logarithm function, ln(x), is defined only for positive real numbers. So, to ensure that f(x) is defined, the argument, x, must be greater than 0.
Now let's look at the function g(x) = f(e^(-f(x))). Here, e represents the mathematical constant approximately equal to 2.71828.
To find the domain of g(x), we need to consider two conditions:
1. The argument of f(x): e^(-f(x)). The exponential function e^(-f(x)) will always be positive, regardless of the value of f(x). This is because negative exponents of positive real numbers result in fractions, and e raised to a positive exponent is always positive. So, we don't have any restrictions on the argument of f(x).
2. The domain of f(x): ln(x). As mentioned earlier, f(x) = ln(x) is only defined for x > 0.
Considering both conditions, we can conclude that the domain of g(x) is the set of all real numbers where x > 0. In interval notation, the domain can be written as (0, ∞).