# Thread: range of the function

1. ## 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 ?

2. Originally Posted by thereddevils
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 $\displaystyle x$.

3. 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: .$\displaystyle {\color{red}(}1,\:2)$

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**

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

. . but that is not in the domain.