f(x) sqrt (x-1)/ (x+1)
Step-by-step eWhen finding domains of functions you start by assuming that all Real Numbers is the domain. And then you look for values for x that must be excluded. There are a variety of things we must not allow to happen:
Zeros in denominators
Negative radicands (the number inside a radical) of even-numbered roots (square roots, 4th roots, 6th roots, etc.)
Zero or negative arguments to logarithms
Other undefined values like tan%28pi%2F2%29
Your function has a denominator and an even-numbered root. So we need to exclude values for x that would make the denominator zero or the radicand negative.
It is pretty easy to see that if x = -1 then the denominator is zero. (If you don't see this, set the denominator equal to zero, x+1 = 0, and solve.) So -1 is a number that must be excluded from the domain.
We must also exclude any x values that make the radicand negative. This is a little trickier than the zero denominator. The radicand is
%28x-1%29%2F%28x%2B1%29
This is a fraction. How do fractions become negative? If the signs of the numerator and denominator are different. We could express the possibilities of positive numerator and negative denominator and negative numerator and positive denominator and solve it. This would look like:
(x-1+%3E+0 and x%2B1+%3C+0) or (x-1+%3C+0 and x%2B1+%3E+0)
But we can simplify this and make the solution easier if we take a moment to figure out which is bigger, x-1 or x+1? After a little thought I hope it is obvious that x+1 will always be bigger than x-1. (After all if we take any random number, x, and add 1 to it won't we always get a bigger number than if we subtract 1 from x?). And if x+1 is always bigger than x-1 and one of them is positive and one of them is negative, won't the bigger number have to be the positive number? With this logic the only possibility is that the numerator is negative and the denominator is positive:
x-1+%3C+0 and x%2B1+%3E+0
This is much easier to solve than the earlier compound inequality. Solving these we get:
x+%3C+1 and x+%3E+-1
These are the values for x that will make the radicand negative. We must exclude these values from the domain.
Combining this with our earlier exclusion of -1, our domain is any numbers less than -1 or any numbers greater than or equal to 1. In interval notation this is:
(-infinity, -1) and [1, infinity)
The range is the set of possible y values. f(x) equals a square root. Square roots, in general, can be zero or any positive number. The domain, which we have already figured out, includes 1. And if x = 1 then the radicand works out to be zero. And the square root of zero is zero. So the range includes zero. (Note that we had to check this. We can't just assume that the radicand could be zero. We must know that the domain includes the number(s) that make the radicand zero.)
And the domain includes numbers that can make the radicand be any positive number. So the range includes all positive numbers. The range of f(x) then, is all non-negative numbers. In interval notation this is:xplanation:
