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Math Help - More meat for the grinder (domain/range)

  1. #1
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    More meat for the grinder (domain/range)

    State the domain and range.

    1. y = 20 - 2x^2

    I know the domain here is all real numbers.

    I am not sure about the range, but am guessing: all real numbers <= 20 Is this correct?

    2. y = 25(-7x-4)^64

    I know the domain here is all real numbers.

    I am not sure about the range, but am guessing all real numbers <= -100 Is this correct?

    3. y = x^2 / (x^2-16)

    How would I do this one?

    Thanks a bunch
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by bobbyboy1111 View Post
    State the domain and range.









    1. y = 20 - 2x^2

    I know the domain here is all real numbers.

    I am not sure about the range, but am guessing: all real numbers <= 20 Is this correct?
    Correct, but guessing is not really a valid answer. For example, you could say, that this function has a graph that is a parabola, open from below, with highest point (0,20). Or you could argue that, since the range of y=2x^2 is [0,+\infty), the range of y=20-2x^2 must be (-\infty,20]

    2. y = 25(-7x-4)^64

    I know the domain here is all real numbers.

    I am not sure about the range, but am guessing all real numbers <= -100 Is this correct?
    No. in fact, 25(-7x-4)^{64}\geq 0 for all x\in\mathbb{R}. - But the correct answer is quite easy to find. Because the range of y=-7x-4 is (-\infty,+\infty) and because the exponent 64 is even (and the factor 25 is > 0), the range of this function must be [0,+\infty)

    3. y = x^2 / (x^2-16)

    How would I do this one?

    Thanks a bunch
    This one is more complicated. First the domain is \mathbb{R}\backslash\{-4,+4\}, because for x=\pm 4 a division by zero would result.
    As to the range: consider first the case that -4<x<+4. In this case, the numerator is always \geq 0 ( =0 for x=0 only), while the denominator is always <0 and tends to 0 from the left (negative side) as x approaches -4 from the right or +4 from the left. So for x\in (-4,+4), this function assumes all values in (-\infty,0].
    Now consider the second case, namely that x<-4 or +4<x. In this case, the numerator x^2 is always positive and larger than the the denominator x^2-16, and hence the function values are \geq 1.
    From y=\frac{x^2}{x^2-16}=\frac{1}{1-\frac{16}{x^2}} we conclude that for x\rightarrow\pm \infty the function values get closer and closer to +1 (from above).
    To sum up: the range is (-\infty,0]\cup (1,+\infty)=\mathbb{R}\backslash (0,1].
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  3. #3
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    #2. - How would I write the answer for this question in the simple format as in question #1 (all real numbers <= 20)? We don't use those sideways figure 8 symbols. So, the domain isn't all real numbers?

    #3. Same here?

    Thanks
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by bobbyboy1111 View Post
    #2. - How would I write the answer for this question in the simple format as in question #1 (all real numbers <= 20)? We don't use those sideways figure 8 symbols.
    Well, yes, if you cannot specify the range in intervall notation (since you are lacking the symbol for "infinity", meaning no upper limit in the case of (a,+\infty) and now lower limit in the case of (-\infty,b), you have to use something like "all real numbers \leq 20".

    #3. Same here?

    Thanks
    Um, yes, but it's going to be a rather clumsy way of expressing the same set. Something like "all real numbers less than or equal to 0 or greater than 1".
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  5. #5
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    Quote Originally Posted by Failure View Post
    Well, yes, if you cannot specify the range in intervall notation (since you are lacking the symbol for "infinity", meaning no upper limit in the case of (a,+\infty) and now lower limit in the case of (-\infty,b), you have to use something like "all real numbers \leq 20".

    Sooo.. this would be written - all real numbers >= 0? That is... for question #2


    Um, yes, but it's going to be a rather clumsy way of expressing the same set. Something like "all real numbers less than or equal to 0 or greater than 1".
    hmm.. is to the second quote... Don't forget to read my question in the first quote in this specific reply.
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by bobbyboy1111 View Post
    hmm.. is to the second quote... Don't forget to read my question in the first quote in this specific reply.
    Sorry, I get confused. Let me just list the answers to your 3 original problems in intervall notation and in plain English:

    1. y=20-2x^2 has domain \mathbb{R}, that is: "all real numbers", and range (-\infty,20], that is: "all real numbers less than or equal to 20".

    2. y=25(-4x-7)^{64} has domain \mathbb{R}, that is: "all real numbers", and range [0,+\infty), that is: "all real numbers greater than or equal to 0" (aka: "all nonnegative real numbers").

    3. y=\frac{x^2}{x^2-16} has domain \mathbb{R}\backslash\{-4,+4\}, that is: "all real numbers except -4 and +4", and range (-\infty,0]\cup (1,+\infty), that is: "all real numbers less than or equal to 0 or greater than 1".

    Hope that helps.
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  7. #7
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    Thanks!!!
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  8. #8
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    One last thing Failure..

    Please re-read my written out range for question #3 and let me know if it is correct = {all real numbers <= 0 or >=1}

    Thanks again
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  9. #9
    Super Member Failure's Avatar
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    Quote Originally Posted by bobbyboy1111 View Post
    One last thing Failure..

    Please re-read my written out range for question #3 and let me know if it is correct = {all real numbers <= 0 or >=1}

    Thanks again
    No, I'm sorry that ({all real numbers <= 0 or >=1}) is wrong. It should be {all real numbers <= 0 or >1}. This is because the equation
    \frac{x^2}{x^2-16}=1 has no solution. So there is no x for which the function assumes the value 1. To see this, just multiply the equation by x^2-16 and subtract x^2: you get 0=-16 which, very clearly, is always false.

    Have a nice day.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Or... x\epsilon\emptyset
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  11. #11
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    Quote Originally Posted by Failure View Post
    No, I'm sorry that ({all real numbers <= 0 or >=1}) is wrong. It should be {all real numbers <= 0 or >1}. This is because the equation
    \frac{x^2}{x^2-16}=1 has no solution. So there is no x for which the function assumes the value 1. To see this, just multiply the equation by x^2-16 and subtract x^2: you get 0=-16 which, very clearly, is always false.

    Have a nice day.
    I thought so, I was just checking
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