Skip to 0 minutes and 11 seconds Hello. 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 seconds What 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 seconds What 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 seconds f 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 seconds No 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 seconds And 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 seconds 1 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 seconds OK, 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 seconds Thank 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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