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# Find the domain of each function.(a) $f(x) = \dfrac{1 - e^{x^2}}{1 - e^{1 - x^2}}$(b) $f(x) = \dfrac{1 + x}{e^{\cos x}}$

## a) $(-\infty,-1) \cup(-1,1) \cup(1,+\infty)$ or $x \neq \pm 1$ depending on your teacherb) $(-\infty,+\infty)$

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### Video Transcript

all right To find the domain of this function, we need to concentrate on the denominator and remember that the denominator cannot equal zero. So we need to figure out what would make that zero. So if one minus C to the power one minus X squared was equal to zero, that would mean that e to the power one minus X squared was equal to one. So how could that possibly be true? So we know that each of the zero is one. So that means that one minus X squared must be equal to zero. So let's solve that for X. That would give us one equals X squared. And if we square rooted both sides, we would get plus or minus one equals X. So that tells us that X cannot be won or ex cannot be negative one, or we will get zero in the denominator of our function. So the domain will be all real numbers from negative infinity to negative. One union negative one toe, one union, one to infinity. That's our way of showing that we're skipping over negative one, and we're skipping over one now. Let's take a look at part B we have the same concept to think about. We don't want the denominator to equal zero. So that means that we need to figure out what makes E to the co sign of X equals zero. Well, what makes it to the anything equal zero? Think about the graph of the each of the X function. It actually never equal zero. It never reaches the height of zero. So this expression will never equal zero either. So that tells us that our domain is all real numbers.

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