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Math Help - Images of sets

  1. #1
    Senior Member Danneedshelp's Avatar
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    Images of sets

    Two questions...

    #1

    Oooopps! I made a big mistake. I had the answer right, I was just looking at the wrong problem in the back of my book. Im sorry about that.

    #2

    Let f:\mathbb{R}\rightarrow\mathbb{R} be the function given by f(x)=x^{2}. Let A=[-3,2] and C=[1,5]. Then f(A\cup{C})=f([-3,5])=[0,25] and f(A)\cup{f(C)}=[0,9]\cup{[1,25]}=[0,25].

    I don't see how f(A)=[0,9] and how [0,25] comes from f(A\cup{B}). Where are those two zeros coming from?

    Thanks for the help,

    Dan
    Last edited by Danneedshelp; August 26th 2009 at 01:12 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    I'll tackle #2

    Quote Originally Posted by Danneedshelp View Post
    #2

    Let f:\mathbb{R}\rightarrow\mathbb{R} be the function given by f(x)=x^{2}. Let A=[-3,2] and C=[1,5]. Then f(A\cup{C})=f([-3,5])=[0,25] and f(A)\cup{f(C)}=[0,9]\cup{[1,25]}=[0,25].

    I don't see how f(A)=[0,9] and how [0,25] comes from f(A\cup{B}). Where are those two zeros coming from?

    Thanks for the help,

    Dan
    Since f(x)=x^2, x\in\mathbb{R}\implies f\!\left(x\right)\in\left[0,\infty\right).

    If A=[-3,2], then f\left(A\right) is the interval of the range over A, which would be \left[0,9\right].

    (You take the smallest and largest value of the range over the interval. Note that f(-3)=9 and f(2)=4, but 0\in\left[-3,2\right]\implies f(0)=0. So the interval for f(A) is \left[0,9\right])

    Similarly, if C=\left[1,5\right], then f\left(C\right) is the interval of the range over C, which is \left[1,25\right].

    So it follows that f\left(A\right)\cup f\left(C\right)=\left[0,9\right]\cup\left[1,25\right]=\left[0,25\right].

    Now, A\cup C=\left[-3,2\right]\cup\left[1,5\right]=\left[-3,5\right].

    Therefore, f\left(A\cup C\right)=f\left(\left[-3,5\right]\right)=\left[0,25\right].
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Chris L T521 View Post

    (You take the smallest and largest value of the range over the interval. Note that f(-3)=9 and f(2)=4, but 0\in\left[-3,2\right]\implies f(0)=0. So the interval for f(A) is \left[0,9\right])
    Is that how I should think about each problem or does that just work for this one?

    edit: I see now, f([-3,2]) is the set of ALL images of elements of [-3,2]. Since -1/2, -2/5, 0 , and so on are in this set, their respective images: 1/4, 4/25, and 0 must be in f([-3,2]). Thus, f([-3,2])=[0,9].

    Thanks again for the help!
    Last edited by Danneedshelp; August 26th 2009 at 01:18 PM.
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