The function f(x) = In(x) has a domain of all real numbers greater than zero and a range of all real numbers. The inverse of this function is f-?(x) = ell. Which conclusion can be drawn by comparing the two functions?

The domain of f(x) is all real numbers and the range is all real numbers.

The domain of f(x) is all real numbers greater than 0 and the range is all real numbers.

The domain of f-1(x) is all real numbers and the range is all real numbers greater than 0.

The domain off-1(x) is all real numbers greater than 0 and the range is all real numbers greater than 0.

The domain of f-1(x) is all real numbers and the range is all real numbers greater than 0.

Step-by-step explanation:

The inverse function of f(x) = ln(x) is:

f-1(x) = exp(x).

And as:

The domain of f(x) is: x ∈ R, x > 0.

The range of f(x) is: x ∈ R

Then for the inverse, we should have the opposite:

The domain of f-1(x) is: x ∈ R  (the exponential function does not have any problems like the logarithm, so you can actually input any real value).

The domain is: x ∈ R, x > 0.l

Because, for example:

e^0 = 1

e^-1 = 1/e > 0.

e^-n = 1/e^n > 0.

e^x is always positive, then the range is the set of all real numbers large than zero.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

the answer on ed. is:

c. "the graph of f(x) has exactly two x-intercepts."

hope i could someone! : )