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Math Help - Exponential functions

  1. #1
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    Exponential functions

    For the function: y = 3(2)^x - 1
    Are these correct?
    Domain: -infinity to infinity
    Range: -1 to infinity
    Y-Asymptote: -1

    I also need to find the intervals of increase/decrease but I am unsure on how to do that.

    And my last question. How do you find the growth rate of 10000(1.06)^x
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  2. #2
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    Quote Originally Posted by olen12 View Post
    For the function: y = 3(2)^x - 1
    Are these correct?
    Domain: -infinity to infinity
    Range: -1 to infinity
    Y-Asymptote: -1

    I also need to find the intervals of increase/decrease but I am unsure on how to do that.
    All of this is correct although with your range you don't include -1 itself. The range should be written as y \in (-1,\infty)


    Quote Originally Posted by olen12 View Post

    I also need to find the intervals of increase/decrease but I am unsure on how to do that.
    I think you mean

    as  x \rightarrow \infty, y \rightarrow \infty

    as  x \rightarrow -\infty, y \rightarrow -1

    Quote Originally Posted by olen12 View Post

    And my last question. How do you find the growth rate of 10000(1.06)^x
    As the exponent has base 1.06 then the growth rate is 6% with an initial value of 10,000

    Hope that helps!
    Last edited by pickslides; May 9th 2009 at 03:11 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by olen12 View Post
    For the function: y = 3(2)^x - 1
    Are these correct?
    Domain: -infinity to infinity
    Range: -1 to infinity
    Y-Asymptote: -1

    I also need to find the intervals of increase/decrease but I am unsure on how to do that.

    And my last question. How do you find the growth rate of 10000(1.06)^x
    The range is -1 \leq y \leq \infty because as you said before there is an asymptote at -1. Otherwise correct.

    The interval of increase or decrease depends on what the value of x is.

    You can find the growth rate by using the binomial expansion (1.06 = 1+0.06) or by plugging in x or using logarithms depending on what you know
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