How to determine the upper and lower bounds for the function $\frac{1}{1+e^x}$? If I plot the equation mentioned on the topic, for $x \in [-10, 10]$, I get the following graph:
$\frac{1}{1+e^x}$ for $x \in [-10,10]$
The plotted graph can also be seen here: [http://m.wolframalpha.com/input/?i=plot+1%2F(1%2Be%5E-x)+x%3D%5B-10,10%5D]
Looking at the plotted graph one can notice that when $x$ reaches $-5$ the function gets really close to $0$ and any number lower than that makes almost no difference. As $x$ reaches $5$, the function gets close to $1$ and any number larger than that makes almost no difference.
My question is: how to determine these boundary values of $-5$ and $5$?
Thanks!
 A: As $x \to \infty$, $e^x \to \infty$ so that $ 1 + e^x \to \infty$ and therefore $$\lim_{x \to \infty}\frac{1}{1+e^x} = 0$$ This is your lower bound since it occurs when the denominator is maximized.  Now try to follow the same reasoning for $x \to -\infty$ to achieve a similar bound.
In case you were actually asking about $\frac{1}{1 + e^{-x}}$, you can follow a similar procedure.  Take $x \to \infty$ and note that $e^{-x} = \frac{1}{e^x} \to 0$ so that $$\lim_{x \to \infty} \frac{1}{1 + e^{-x}} = 1$$
A: Your graph will never reach $0$ or $1$. (These are the asymptotes of this graph.)
You can verify they are never reached by trying to solve the function equal to these two values:
Equal to zero
$$0=\frac{1}{1+e^{-x}}$$
cross multiply
$$0=1$$
No solution
Equal to one
$$1=\frac{1}{1+e^{-x}}$$
cross multiply
$$1+e^{-x}=1$$
$$e^{-x}=0$$
Again no solution.
A: Let $x_1 , x_2 \in [-10,10] $ be arbitrary.
For $x_1<x_2$ ,  
Observe that 
$$e^{x_1}<e^{x_2}$$ since $e^x$ is a strictly increasing function. 
$$1+e^{-x_1}>1+e^{-x_2}$$
$$1+e^{x_1}<1+e^{x_2}$$
$$\frac{1}{1+e^{x_1}}>\frac{1}{1+e^{x_2}}$$
Thus $$f(x)=\frac{1}{1+e^x}$$ is a strictly decreasing function.
So for $x \in [-10,10]$ ,
$$f(x)_{min}=f(x)=\frac{1}{1+e^{10}} >0 $$ and
$$f(x)_{max}=f(x)=\frac{e^{10}}{1+e^{10}} <1$$
You can show that $f_{min}$ and $f_{max}$ are its supremum and  infimum  . Hence upper bound and lower bounds.
A: Perhaps the most standard and general way tosolve this is to approach it like you would approach an "Absolute Min/Max" problem. That is, set the derivative equal to zero to find local/relative max/min, and evaluate the function at the endpoints of the interval ("evaluate" at $\pm \infty$ in this case)
$$f(x) = \dfrac{1}{1+e^x}$$
$$f(x) = (1+e^x)^{-1}$$
$$f'(x) = -e^x (1+e^x)^{-2}$$
$$f'(x) = - \dfrac {e^x}{(1+e^x)^2}$$
$$0 = - \dfrac {e^x}{(1+e^x)^2}$$
A fraction is be equal to zero if and only if the numerator is zero. Since $e^x$ is never zero for any $x$, this has no solution. Therefore, there are no relative extrema. 
Now just combine Yushwuth's limit argument with this answer, and show that for $1$ $x$, $f(x)$ is between $0$ and $1$ ($f(0) = \dfrac 12$ for example) and you're done.
A: Okay, so after reading Ian's and Yushwuth's answers and comments I realized that what I was trying to accomplish was simple. I apologize for not being 100% clear on what I was trying to achieve in my question, but I hope the comments made it clear.
I am trying to use this function as a spline interpolation to draw a color gradient. So the idea is that the function will give me values between 0% and 100%. Considering this, what needs to be done to figure out what are the values for $x$ that approach that, I just need to calculate the following two equations:
$\frac{1}{1+e^x} = 0.99$
$x \approx 4.59$
and
$\frac{1}{1+e^x} = 0.01$
$x \approx -4.59$
These will give me the values for $x$ that allow me the range 1% to 99%.
Thanks for all other replies! 
