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Math Help - Function limits-f and g

  1. #1
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    Post Function limits-f and g

    Let h(x)=2f(x) - g(x), where f and g are functions satisfying:

    \lim_{x\to3}f(x)=-2,

    \lim_{x\to3}g(x)=4,

    \lim_{x\to7}f(x)=6,

    \lim_{x\to7}g(x)=-5.

    Find the following:

    a. \lim_{x\to3}h(x)

    b. \lim_{x\to7}(g(x)/h(x))

    c. \lim_{x\to3}f(x)*g(x+4)


    Here are the answers I have..but I'm not sure if I'm doing the problems correctly or if my answers are correct. An explanation would be great!
    a. -8
    b. -5/17
    c. -16
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  2. #2
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    Quote Originally Posted by live_laugh_luv27 View Post
    Let h(x)=2f(x) - g(x), where f and g are functions satisfying:

    \lim_{x\to3}f(x)=-2,

    \lim_{x\to3}g(x)=4,

    \lim_{x\to7}f(x)=6,

    \lim_{x\to7}g(x)=-5.

    Find the following:

    a. \lim_{x\to3}h(x)

    b. \lim_{x\to7}(g(x)/h(x))

    c. \lim_{x\to3}f(x)*g(x+4)


    Here are the answers I have..but I'm not sure if I'm doing the problems correctly or if my answers are correct. An explanation would be great!
    a. -8
    b. -5/17
    c. -16

    a) Correct.

    b) Correct.

    c) Incorrect.


    Note that when x \to 3, g(x + 4) \to g(7) = \lim_{x \to 7}g(x).


    So \lim_{x \to 3}f(x)\cdot g(x + 4) = \lim_{x \to 3}f(x)\cdot\lim_{x \to 3}g(x + 4)

     = \lim_{x \to 3}f(x)\cdot\lim_{x \to 7}g(x)

     = -2\cdot(-5)

     = 10.
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  3. #3
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    Post

    Also, what additional information would you need to know about g in order to prove that there is a value between 3 and 7, such that g(x)=0?
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  4. #4
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    Quote Originally Posted by live_laugh_luv27 View Post
    Also, what additional information would you need to know about g in order to prove that there is a value between 3 and 7, such that g(x)=0?
    Isn't it obvious that if \lim_{x \to 3}g(x) = 4 and \lim_{x \to 7}g(x) = -5, then the function must cross the x-axis somewhere in between in order to allow for the change in sign?
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  5. #5
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    yeah it is, now that I look at it. So by using IVT, you need to know the information that g(x) crosses the x axis at some point, which would make g(x) = 0?
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  6. #6
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    Quote Originally Posted by live_laugh_luv27 View Post
    yeah it is, now that I look at it. So by using IVT, you need to know the information that g(x) crosses the x axis at some point, which would make g(x) = 0?
    Correct.

    Edit: This is assuming, of course, that the function g(x) is continuous on the interval x\in (3, 7).
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