Find the domain of each function

[gr8joel]

f(x)= 1/1+e^x

f(x)=1/1-e^x

I have never encountered such a problem before. It is in my Calculus Early Transcendentals 6th E. book so I figured I posed in the right section. Honestly I have no idea how I would go about even starting to find the domain of such a problem. I graphed it on a graphing calculator which then gave me an answer for the first problem. I also graphed the second equation which also gave me an answer for the second problem. They are as follows:

First problem- Domain= (-infinite,+infinite)
Second problem- Domain= (-infinite,0)U(0,+infinite)

I have cross-referenced the answers in the back of my textbook and these are indeed the answers to my problem.

I just want to know how would one go about figuring out the domain of such a problem without a graphing calculator?

Thanks in advance

 

[mmm4444bot]

f(x) = 1/1+e^x

f(x) =1/1-e^x

I have never encountered such a problem before. It is in my Calculus Early Transcendentals 6th E. book

Did you study pre-calculus? Determining domains of functions is a precalculus topic. (In a calculus text, domain exercises are review.)

It seems that you have learned the definition of domain because you were able to determine the answers graphically.

Here's one way to think about it. The domain is the set of all valid input values for the function (i.e., all values of x such that f(x) evaluates to a Real number).


f(x) = 1/(1 + e^x)

A fraction defines this function, yes? (The way that you typed it above actually means something else. See my comment at the end of this post about grouping symbols.)

In the Real number system, we cannot divide by zero. Thus, the fraction 1/0 is not defined.

The fraction 1/(1 + e^x) will be a Real number, so long as the denominator is not zero. In other words, the expression 1 + e^x cannot equal zero. Any value of x that leads to zero is invalid and not in the function's domain.

All other Real numbers make up the domain.

However, when we consider what values of x might lead to division by zero in this function, we realize that there are none. So, the domain is all Real numbers because no value of x leads to division by zero.


Here's how we reason it out.

If the expression 1 + e^x evaluates to zero, then the term e^x must be -1, yes? But the term e^x is positive for all values of x, so e^x will never be -1.

We know that e^x is always greater than zero because we learn about the exponential function y = e^x before getting exercises that contain it.


f(x) = 1/(1 - e^x)

With this function, the domain consideration is the same. The fraction is defined for all values of x that do not lead to zero in the denominator. In this function, the denominator becomes zero when the term e^x equals 1.

What value of x causes e^x to be 1? That value cannot be in the domain.

Again, we know exactly which value this is because we've previously studied elementary exponential functions and learned that in each case their y-intercept is 1. In other words, raising a base to the power of zero always equals 1.


Here is a note about typing grouping symbols.

You typed 1/1 + e^x. The Order of Operations says "do division before addition". So your typing actually means f(x) = 1 + e^x.

What you mean is 1/(1 + e^x). Here, the grouping symbols change the order so that the addition is done first, and we have a fraction (1 in the numerator and 1+e^x in the denominator).

When you type fractions with a keyboard in the future, please be sure to include the grouping symbols!


If there is anything in my post the you do not understand, please ask specific questions. Cheers

 

[gr8joel]

yes I have studied pre-calculus. sorry for not grouping it the way it should be. I will correctly group my symbols next time. By the way, I would like to say that you explained yourself thoroughly and I am quite satisfied with the way you explained it. Thank you very much.