- f(x) = 1/(2 + ex)
For the domain, we need to find the set of values that we can possibly input for x and the function would work. In most cases, it is easiest to find the values wherein the function does not work, so let's do that. In a fraction like this, we know that the denominator cannot equal 0, because we cannot divide by 0. So let's see if there's any x value that would cause that:
2 + ex = 0
ex = -2
Now a positive number to any exponent can never equal a negative number, so we know there is no value of x that can cause the denominator to be 0.
Thus the domain is all real numbers.
The range is the range of values that we can expect to get out of this function. So looking back at the equation, we see that the only part that can change is the e^x. The value of e^x can be anywhere from an infinitely small positive number (when x approaches negative infinity) to an infinitely large positive number (when x approaches positive infinity).
When e^x is a small positive number, approaching but not quite equal to 0, our fraction approaches 1/2 (but does not touch it). So we know that f(x) has an open bound at 1/2.
When ex is an infinitely large positive number, our denominator becomes infinitely large, and so the value of our fraction approaches 0 (but does not touch it). So we know that f(x) has an open bound at 0.
Thus, our range is the real numbers, with 0 < y < 1/2.
For the rest of the problems, you just follow a similar thought process. For domain, what x can we input in the function that will not allow the function to compute? Those values are excluded from the domain. For range, considering all the values that we can input into the function, what range of outputs can we expect to see come out? That will be our range.
