33 Replies to “MitPY 5: Implied Domain and Range”

  1. For example, finding the implied domain of the function y=-√(9-x^2)

    An example in the textbook is f(x) =√(2x-5) where it states that for f(x) to be defined they need 2x-5 ≥ 0 but it doesn’t say why, which is what I don’t understand. (Marty edit: brackets added.)

    1. Thanks, Michaela. I’ve added brackets, so the functions are what I think you intended. For your second example, how do you (Michaela) think about what the domain is? That is, what does the word “domain” mean to you?

        1. The implied domain would be positive real numbers I think maybe. Because you can’t square root a negative

          1. Hi, Michaela, basically that’s right, for the reason you give, although 0 is also ok. So the domain is nonnegative real numbers.

        1. Hi, Michaela. In practice that’s how you often work out what the domain is. (You may notice some of the commenters here are arguing about how and how much to focus on this top number and bottom number idea.) But, it’s not really getting to what domain means, the underlying idea.

          The domain of a function f(x) just means the values of x that we’re allowed to plug into the function. For example, as you correctly pointed out in your comment below, the domain of f(x) = 7 – x will be all real numbers. Why? Because, we can plug in any value of x that we like, and no problems will occur.

          Just two quick howevers:

          1) Domains are officially sets of numbers. So the domain of f(x) = 7 – x will be the set of all real numbers.

          2) What we’re talking about here is the implied (or maximal) domain, which is what you asked about. That means we’re allowing all x that we conceivably can allow to be in the domain. There are other notions of domain, where someone might so “No, I don’t like the number 3. So, x = 3 is not in the domain. I’m just not allowing it.” (Why someone would do this can wait until later!)

          1. ohhhh right I understand that. So the domain of f(x)=√x cannot be a negative number because then there would be an error which is why the domain is non negative numbers

            1. Exactly! The implied domain is (the set of) all numbers that don’t result in errors.

              (There’s fancier language for “error”, but your sense is spot on.)

            2. Hi Michaela, it seems you’re pretty right with implied domains (but we haven’t talked about ranges yet). Just as a check, how would go about figuring out the domain of f(x) = \sqrt{9-x^2}?

              If that example works out ok, I’ll update the post with a summary. Then, if you wish, you can ask about corresponding ranges of functions.

              1. I would do 9-x^2≤0
                and then that comes out as x≤3

                But when I look in the answers in the book, the way that they write the answer is written as f: [-1,2] -> R, f(x) = 5x+6

                1. Hi Michaela.

                  If the question is to find the maximal domain of f(x) = 9 - x^2, then your inequality 9 - x^2 \leq 0 is correct. Good work!

                  But your solution to the inequation is only half right ….

                  Now, this might seem irrelevant, but humour me ….. I want you to try and draw the graph of y = 9 - x^2. Don’t worry too much about labelling the coordinates of the turning point, but I really want you to label the x-intercepts.

                  Please try to do this. It’s a very important part of solving the inequality. If you’re having trouble drawing this graph, finding the x-intercepts etc. please let us know.

                  As for what the answers in your book say, I don’t see a connection to finding the maximal domain of f(x) = 9 - x^2 …. Are you checking the correct answer at the back of the book?

                  1. To graph it I re-arranged it to be y=-x^2 + 9 which I know is a parabola translated up the y axis 9 units and it gave me x intercepts -3 and 3

                    1. Good work (but I hope it’s an ‘upside down’ parabola ….). Now do you see that the inequality 9 - x^2 \geq 0 means the same thing as y \geq 0?

                      So, looking at the graph you drew, can you see the values of x that make y \geq 0 true? Those values of x are your answer!

                2. Hi, Michaela. I assume that you, and JF, mean 9 – x^2 ≥ 0, which is correct. But then you have to think more carefully about which x will make that true. JF is giving you one approach to doing that.

                  I’m not sure I understand your comment about the book answer. If you’re just worried about the format of their answer, with all the weird notation, that’s easy to sort out later.

    2. Hi Michaela and welcome to Marty’s Help in the Plague blogs.
      Well done for doing some work during the holidays (I guess there’s nothing much else to do, is there …lol!)

      Let’s say you have \sqrt{a} = b where a and b are real numbers (real numbers are the only type of numbers used in Maths Methods). This means that a = b^2. For example, \sqrt{9} = 3 means that 3^2 = 9. So far, so good …?

      Now, can we agree that if a < 0 (that is, a is negative) and a = b^2, then b^2 < 0. For example, if a = -1, then b^2 = -1. Can you find a real number b such that b^2 = -1 ….? (I hope not!). So there is no real number b such that \sqrt{-1} = b.

      I hope you’re still with me ….

      So, in Maths Methods when you have \sqrt{\text{stuff}}, the stuff is not allowed to be negative. For \sqrt{\text{stuff}} to be defined in Maths Methods, the stuff has to be greater than or equal to zero.

      In your textbook example, the stuff is 2x-5, so for the square root to be defined you need 2x-5 \geq 0.

      I hope this makes sense for you.

      Best wishes with your studies.

    3. Hi, Michaela. Are the commenters’ suggestions making some sense to you? I’ll update the post with a summary at some point, but first I want to see if you’re actually getting some help from the help.

      Feel free to ask follow-up questions (to anyone or in general, suggest what is (or is not) helpful.

  2. At this stage in Year 11, there are two main cases where the maximal domain of f(x) will be something other than R:

    1) \sqrt{x} requires x \geq 0. No negative number is the square of a real number (the square of a non-negative number is either zero or positive) hence no negative number has any (real) square roots.

    2) \frac{1}{x} requires x \neq 0. Division by zero is undefined in the real numbers.

    So, for the sorts of cases you will see at this stage:

    To find the maximal domain of something like \frac{1}{f(x)} (where f(x) is a “rule” involving x), solve f(x) = 0 for x, and the maximal domain will be all real numbers except for those solutions.

    To find the maximal domain of something like \sqrt{f(x)}, solve f(x) ≥ 0 for x, and this set of solutions is the maximal domain.

    To find the range, at this stage the best method is to do a quick sketch of the graph of y=f(x) and the range is the minimum y coordinate of a point on the graph to the maximum y coordinate of a point on the graph. When sketching, the key features you need to find the location of are: (i) any turning points (such as on a parabola), (ii) any endpoints (such as on a graph of a square root function, (iii) any asymptotes (such as for a hyperbola or truncus). You don’t need to find intercepts (unless they are one of these points). And of course you don’t need to sketch the graph if you can visualise it, but at least make sure you can reliably find the location of those key features.

    1. Thanks to readers for being polite enough to overlook (or charitably interpret) my error in 1). I meant to say that the square of a real number is either zero or positive.

  3. Regarding the statements in the last paragraph of SRK, a clarification may be in order: while checking the existence and location of turning points of functions is surely an important exercise generally, there is no connection between turning points and the range of an “arbitrary” function. Parabolas, where the turning point is also the minimum or maximum, are surely important functions, especially at Year 11. Nevertheless, it is still good (IMHO) to keep in mind that they are but one kind of animal in the much larger zoo of functions for which the said connection does not exist.

    1. Thanks Christian, I agree there’s always a delicate balancing act between, on one hand, giving students concrete cases to learn an important idea and, on the other, discouraging incorrect generalisations.

      What I’ve found, though, is that if asked to find the range of a function on a bounded interval (where a student’s “repertoire” of functions consists of f(x)=x^n,\;n \in { -2,-1,0,\frac{1}{2},2 } the most common mistake at this stage is to find the value of the function at endpoints and ignore the location / existence of turning points. So while, in general, there is no strong connection (apart from differentiable functions on closed bounded intervals), finding turning points on parabolas helps students to appreciate that the max / min may not occur at endpoints, and hence this needs to be checked.

      Edit: Does anyone know how to typeset curly brackets? I tried “forward-slash {” and “forward-slash }”, but it is not showing…

      1. in LaTeX, try {}, as braces are used for semantic grouping, and the backslash escapes them.

        EDIT: Supposed to read backslash opening curly bracket backslash closing curly bracket

  4. Thank you SRK for your very sensible comments. An important correction to my own comment from yesterday: a parabola \latex y=ax^2+by+c has constant second derivative \latex 2a and hence no turning point at all, at least if defined in the standard way. Sorry for the oversight.

    1. Have I missed something here or does the constant second derivative not actually matter for a parabola (also, should it be “+bx” not “+by”? Surely it is the sign of the second derivative that we are interested in?

      Also, when you say “defined in the standard way”… is there another way to define a turning point other than a point at which the sign of the gradient changes (smoothly)?

      Forgive if this has already been covered earlier and I just missed the important parts.

      1. Thank you for your message. First, the easy bit. Yes the correct formula for the parabola (one that is opened in direction of the \latex y axis) is \latex{y=ax^2+bx+c}. Sorry for the typo.

        Your other comment showed me (at last!) that I piled a language issue (as an ESL speaker who, as a minor excuse, never taught basic calculus in English) on an oversight, even though these issues appeared to have (sort of) canceled out each other. I wrote under the erroneous assumption that a “turning point” was actually synonymous to “point of inflection”, as the German word for the latter term, namely “Wendepunkt”, has the same root as the verb “wenden” = “to turn”. This should clear up the last question in your first paragraph (to which the answer is “yes”, with the small addendum that we may not want, or need, to compute the second derivative if we have some local information on the first near where it is zero); and it may serve as an answer to the one in the second paragraph, too.

        I truly apologize for this confusion and hope that other readers, including Michaela, are accommodating.

  5. Hi Terry, sorry just noticed your comment was in the trash for some reason. (In fact, I’m not thrilled with your picture, but I didn’t trash the comment, at least not intentionally.)

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