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

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

    Thanks for the help,

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

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

    Thanks for the help,

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

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

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

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

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

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

    Therefore, $\displaystyle f\left(A\cup C\right)=f\left(\left[-3,5\right]\right)=\left[0,25\right]$.
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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 $\displaystyle f(-3)=9$ and $\displaystyle f(2)=4$, but $\displaystyle 0\in\left[-3,2\right]\implies f(0)=0$. So the interval for $\displaystyle f(A)$ is $\displaystyle \left[0,9\right]$)
    Is that how I should think about each problem or does that just work for this one?

    edit: I see now, $\displaystyle f([-3,2])$ is the set of ALL images of elements of $\displaystyle [-3,2]$. 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 $\displaystyle f([-3,2])$. Thus, $\displaystyle f([-3,2])=[0,9]$.

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