Skip to 0 minutes and 11 secondsHello. I am Alberto. Nice to meet you in this FutureLearn course. Welcome to the steps in practice of week two. Let us start with the first exercise of the function concept in practice-- step. We have to compute the domain and the range of this function. Let us start to compute the domain.

Skip to 0 minutes and 37 secondsWhat it means to compute the domain of the function-- I am asking myself, which is the biggest subset of real numbers where this function is well-defined? It's clear that here, we have a problem with this denominator. As you know, the denominator cannot be equal to 0. Therefore, we have to impose that x square minus one is different from 0, that means x is different from plus and minus 1. For all the other real numbers, this function is well-defined. Therefore, the domain of f is equal to the set of real numbers, minus or plus, and then minus 1. And now, let us consider the range of our function.

Skip to 1 minute and 35 secondsWhat it means to compute the range of a function-- OK, given y in R, any real number, I am asking myself if it is possible to reach y throughout the function, f. That means given y, is it possible to find x in R such that y is equal to f of x?

Skip to 2 minutes and 10 secondsf of x is equal to 1 over x square minus 1. Therefore, the problem is given y, the equation y equal to 1 over x squared minus 1, admits some solutions in x or not. OK, x has to belong to the domain of f. Therefore, x is different from plus and minus 1. Then this equation is equivalent to the equation x squared minus 1 times y equals 1 that we can get from this one, just multiplying on both sides by x square minus 1. And now, I immediately realize, for example, that if y is equal to 0, this equation has no solutions.

Skip to 3 minutes and 6 secondsNo solutions. What it means-- it means that 0 does not belong to the range of our function f. Now assume that y is different from 0. Then we can divide on both sides by y. And this equation is equivalent to x square minus 1 is equal to 1 over y. That is x squared equals 1 over y plus 1, which is equal to 1 plus y over y. Now we have to consider this new equation. OK since x to the square is always greater or equal than 0 for any real number, then we can solve this equation if and only if 1 plus y over y is greater or equal than 0.

Skip to 4 minutes and 21 secondsAnd we find the condition 1 plus y over y greater or equal than 0. OK, let us try to understand better what this condition means. We have the following possibilities. First possibility-- 1 plus y over y equal to 0. Second possibility-- 1 plus y greater than 0, and y greater than 0. And the third possibility--

Skip to 5 minutes and 7 seconds1 plus y less than 0. And y less than 0. OK the first one, the first one means exactly y equal to minus 1.

Skip to 5 minutes and 27 secondsOK, the first is y greater than minus 1. And y greater than 0. Then this is equivalent to y greater than 0. And the third condition-- this means y less than minus 1, and y less than 0. This is equivalent to y less than minus 1. OK, then if y is equal to minus 1, or y is greater than 0, or y is less than minus 1-- in all these cases, our equation admits a solution. And therefore we have elements of the range of f. Then we can conclude that the range of f is obtained joining all the real numbers, which are less or equal than minus 1 and all the real numbers that are strictly greater than 0.

Skip to 6 minutes and 53 secondsThank you for your attention.

# The function concept in practice

The following exercises are solved in this step.

We invite you to try to solve them **before** watching the video.

In any case, you will find below a PDF file with the solutions.

### Exercise 1.

Compute the domain and try to figure out the range of the function \[f(x)=\frac 1{x^2-1}.\]

### Exercise 2. [Solved only in the PDF file]

Compute the domain and try to figure out the range of the function \[g(x)=\frac{x+3}{x-3}.\]

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