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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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: $$\boxed{\;\mbox{range} (f)=\left(0,1/2\right]\;}$$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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! Answers


I 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 \(\displaystyle \displaystyle \begin{align*} e^x > 0 \end{align*}\) for all x, so is \(\displaystyle \displaystyle \begin{align*} 1 + 2e^x \end{align*}\). So the domain is \(\displaystyle \displaystyle \begin{align*} \mathbf{R} \end{align*}\).

As for the range, for starters

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

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

Editing...