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Domain and Range of this function

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the domain and range of the function [itex] y = \frac{2}{x^2-4}[/itex]



    3. The attempt at a solution
    I'm lost, and I noticed it seems like a difference of squares so what I did is y = [itex]\frac{2}{(x-2)(x+2)}[/itex]. So the x value of vertex is 0. And i subbed this into the equation to get y = 2/-4
    y = -1/2.

    I'm guessing Domain is {x[itex]\in[/itex][itex]\Re[/itex]}
    and Range is {y[itex]\in[/itex][itex]\Re[/itex] | y [itex]\geq[/itex] [itex]\frac{-1}{2}[/itex]

    Is this right?
     
  2. jcsd
  3. Mar 19, 2013 #2
    Your domain is all numbers that you can put in for x which will give you a value for y. So in the case of fractions, you have to find out when the denominator is 0. x values that result in the denominator being 0 are not included in your domain, since its not possible to divide by 0.
     
  4. Mar 19, 2013 #3

    SammyS

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    Both the domain & range you have given are incorrect.

    What is 2 divided by zero ?
     
  5. Mar 19, 2013 #4
    Ah so when x = ±2, it is undefined.

    So... domain: {x∈R | x ≠ ±2}

    What do I do for range? when i plug it in i get undefined?
     
  6. Mar 19, 2013 #5
    There's probably a better way... but I usually start by making a table of values to try and get a feel for the behavior of the graph. When x is small, where is y? As x gets larger, what does y do? Especially check near the points which were not in the domain.

    *note* check near +/-2, not exactly at those values.
     
  7. Mar 19, 2013 #6

    SammyS

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    Yes. When x = ± 2, y is undefined.

    Finding the range is trickier.

    Have you studied graphs of rational functions?
     
  8. Mar 20, 2013 #7

    HallsofIvy

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    One general method of finding is to solve the equation for y. Since it is typically easier to find "domain" than "range", reversing x and y makes it a little easire.

    However, in this case, an important rule is that a fraction is 0 if and only if its denominator is 0.
     
  9. Mar 20, 2013 #8
    @HallsofIvy: I guess you meant "if and only if the NUMERATOR is zero, and the DENOMINATOR is nonzero" (both these conditions must be satisfied)? :)

    @OP: As for the range, I think the general approach is to draw the graph of the function, and take all possible values of y (images of x). In our problem, the first step (after finding the domain) that you should execute is to find the derivative of y (denoted as y') using the quotient rule. Then you proceed to find the possible critical points of the function, i.e. the values of x which for which y' (y prime) = 0 (or y' is not defined). The idea of this is for us to determine the intervals on which the function is increasing or decreasing (specifically if y'<0 => decreasing; y'>0 => increasing). This is arguably the most crucial step in curve sketching. One thing to keep in mind is that the critical points that we have found above are the "candidates" for local/absolute mimima/extrema (we don't know yet, we have to verify whether they are by plugging these values into our function).

    Additional steps include finding the vertical/horizontal/slant asymptotes (if there are any), but these are peripheral in finding the RANGE of our function.

    Warning: 0 is NOT in the range of the function, since there is no value of x for which y=0!

    Btw, I have a "memory trick" which I believe would help you understand the fundamentals of curve sketching: "X-dom, Y-range," meaning that whenever you're asked to find the domain, think of the x-axis and ask yourself "what values of x don't make sense?" (usually for problems like this, the condition for x is that x must be different from some number, and the domain is simply R\{that number}). Conversely, when you're asked to find the range, think of the y-asis, and try to determine, from the graph, what values of y can the function take?

    I hope that helps :)
     
  10. Mar 20, 2013 #9

    eumyang

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    But this thread is in the Precalculus subforum. Unless the OP is currently taking Calculus and is reviewing Precalculus material on the side, this procedure is not going to help him/her.
     
  11. Mar 20, 2013 #10
    Yeah you're right. My guess is that the OP is currently covering this material in class, and the homework is designed by his/her professor to supplement the lectures. Otherwise I cannot think of an alternative approach to this problem without using the derivative. The tabular method is very time-consuming and thus needs the help of computer applications to demonstrate. Also, it does not provide a complete picture if one executes it manually.
     
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