Results 1 to 3 of 3

Thread: range of the function

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    range of the function

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

    Attempt :

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

    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 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by thereddevils View Post
    Find the range of the function , $\displaystyle f(x)=-e^{-2x}+2e^{-x}+1$ , x is real and x>0

    Attempt :

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

    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 ?
    $\displaystyle f(x) = -e^{-2x} + 2e^{-x} + 1$

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

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

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

    $\displaystyle -X^2 + 2X + 1$

    $\displaystyle = -(X^2 - 2X - 1)$

    $\displaystyle = -[X^2 - 2X + (-1)^2 - (-1)^2 - 1]$

    $\displaystyle = -[(X - 1)^2 - 2]$

    $\displaystyle = -(X - 1)^2 + 2$.

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

    And since $\displaystyle X$ is defined for all $\displaystyle x$, that means that $\displaystyle f(x) < 2$ for all $\displaystyle x$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, thereddevils!

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


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


    Attempt :

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

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

    The range would be [1,2)

    I agree with your upper limit, 2.


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

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


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

    . . but never equals 1. **

    So, 1 is not included in the range.


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


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

    **

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

    . . but that is not in the domain.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The range of this function?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 17th 2010, 12:42 PM
  2. Range of a function?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Dec 18th 2009, 01:00 PM
  3. Range of function
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Jun 6th 2009, 04:40 AM
  4. Need the range of this function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 6th 2008, 06:40 PM
  5. range of function?
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: Jan 8th 2008, 02:00 PM

Search Tags


/mathhelpforum @mathhelpforum