MHB Myles's question at Yahoo Answers (Domain and range)

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The function f(x) = (e^x)/(1+2e^x) has a domain of all real numbers, as the denominator is always positive. The range is determined to be (0, 1/2] since f(x) approaches 0 as x approaches both positive and negative infinity, and reaches a maximum of 1/2 at x = 0. The function is positive for all x, confirming that the range is limited to positive values. The analysis includes finding critical points and applying the Intermediate Value Theorem. Overall, the domain and range have been clearly established through mathematical reasoning.
Fernando Revilla
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Here is the question:

Please show all the steps to how you get the answers. I would like to learn how you got to the final result.

Here is a link to the question:

Find the domain and range of f(x)= (e^x)/(1+2e^x)? - Yahoo! AnswersI have posted a link there to this topic so the OP can find my response.
 
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Hello Myles,

For all $x\in\mathbb{R}$ there exist $e^x$ and $e^{2x}$. Besides, $1+e^{2x}>0$ so the quotient exists for all $x\in \mathbb{R}$. as a consequence $$\boxed{\;\mbox{Dom }(f)=\mathbb{R}\;}$$ Clearly $f(x)>0$ in $\mathbb{R}$ and we are going to study the variation of $f$: $$\lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}\frac{e^x}{1+e^{2x}}=\lim_{x\to +\infty}\frac{e^{-x}}{e^{-2x}+1}=\frac{0}{0+1}=0\\\lim_{x\to -\infty}f(x)=\lim_{x\to -\infty}\frac{e^x}{1+e^{2x}}=\frac{0}{1+0}=0$$ This means that $\mbox{range }(f)\subset (0,+\infty)$. Let's find the singular points: $$f'(x)=\ldots=\dfrac{e^x(1-e^{2x})}{(1+e^{2x})^2}=0\Leftrightarrow1-e^{2x}=0\Leftrightarrow x=0$$ If $x<0$, then $f'(x)>0$. If $x>0$, then $f'(x)<0$, so there is an absolute maximum: $f(0)=1/2$. According to the Intermediate Value Theorem for continuous functions, we conclude: $$\boxed{\;\mbox{range} (f)=\left(0,1/2\right]\;}$$
 
Fernando Revilla said:
Here is the question:
Here is a link to the question:

Find the domain and range of f(x)= (e^x)/(1+2e^x)? - Yahoo! AnswersI have posted a link there to this topic so the OP can find my response.

Since it's a fraction, the denominator can never be 0. But since [math]\displaystyle \begin{align*} e^x > 0 \end{align*}[/math] for all x, so is [math]\displaystyle \begin{align*} 1 + 2e^x \end{align*}[/math]. So the domain is [math]\displaystyle \begin{align*} \mathbf{R} \end{align*}[/math].

As for the range, for starters

[math]\displaystyle \begin{align*} e^x > 0 \end{align*}[/math] and [math]\displaystyle \begin{align*} 1 + 2e^x > 0 \end{align*}[/math], so their quotient is also positive. As for the rest, it might be easier to write this in a more standard form...

[math]\displaystyle \begin{align*} y &= \frac{e^x}{1 + 2e^x} \\ &= \frac{1}{2} \left( \frac{2e^x}{1 + 2e^x} \right) \\ &= \frac{1}{2}\left( \frac{1 + 2e^x - 1}{1 + 2e^x} \right) \\ &= \frac{1}{2} \left( 1 - \frac{1}{1 + 2e^x} \right) \\ &= \frac{1}{2} - \frac{1}{2 + 4e^x} \end{align*}[/math]

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