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Math Help - range of the function

  1. #1
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    range of the function

    Find the range of the function , f(x)=-e^{-2x}+2e^{-x}+1 , x is real and x>0

    Attempt :

    f(x)=-\frac{1}{e^{2x}}+\frac{2}{e^x}+1

    so when x=0 , f(x)=2

    and when x approaches infinity , f(x) converges to 1 .

    the range would be [1,2)

    is this acceptable ?
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Find the range of the function , f(x)=-e^{-2x}+2e^{-x}+1 , x is real and x>0

    Attempt :

    f(x)=-\frac{1}{e^{2x}}+\frac{2}{e^x}+1

    so when x=0 , f(x)=2

    and when x approaches infinity , f(x) converges to 1 .

    the range would be [1,2)

    is this acceptable ?
    f(x) = -e^{-2x} + 2e^{-x} + 1

     = -(e^{-x})^2 + 2e^{-x} + 1.

    This is a quadratic in e^{-x}.

    So let X = e^{-x} and the equation becomes

    -X^2 + 2X + 1

     = -(X^2 - 2X - 1)

     = -[X^2 - 2X + (-1)^2 - (-1)^2 - 1]

     = -[(X - 1)^2 - 2]

     = -(X - 1)^2 + 2.

    It should be clear that this is a quadratic with a maximum at (1, 2). So f(X) < 2 for all X.

    And since X is defined for all x, that means that f(x) < 2 for all x.
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  3. #3
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    Hello, thereddevils!

    Good job! . . . I'd make one change.


    Find the range of the function: . f(x)\:=\:-e^{-2x}+2e^{-x}+1,\;x\text{ is real and }x>0


    Attempt :

    f(x)\:=\:-\frac{1}{e^{2x}}+\frac{2}{e^x}+1

    So when x=0.\:f(x)=2
    . . and when x \to\infty,\;f(x) converges to 1 .

    The range would be [1,2)

    I agree with your upper limit, 2.


    Since x > 0, the function is never quite equal to 2.

    . . So, 2 is not included in the range.


    As x\to\infty, the function approaches 1,

    . . but never equals 1. **

    So, 1 is not included in the range.


    The range is: . {\color{red}(}1,\:2)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    I found that f(x) equals 1 when x\,=\,-\ln2

    . . but that is not in the domain.

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