Finding range for $\frac{1}{\exp(x^2)+3}$ What is the range of $$\large\frac{1}{e^{x^2}+3}$$
I know that the answer is $\dfrac{1}{4}\ge h(x)\gt0$, but how do I show it

 A: If you know Calculus, there's a theorem that can help you.
Let $f$ be a (piecewise) differentiable real function with $D_xf$ its derivative WRT $x$.
If $f$ has a maximum or minimum at $a$, then $D_xf(a)= 0$ or $D_xf(a)$ does not exist.
There's only one point where $D_xf$ qualifies but this is a necessary but not sufficient condition (Google necessary and sufficient conditions if those phrases are unfamiliar).
You need to check $f(a - \epsilon)$ and $f(a + \epsilon)$ for a small positive real number $\epsilon$ to see if it's a max, min, or neither. If those two numbers are smaller than $f(a)$, you have your maximum. 
That proves $h(x) \le 1/4$. It also proves that $h$ has no other minima or maxima. Can you think of a way to prove that $h(x) > 0$?
Once you prove that $0 < h(x) \le 1/4$, there's a theorem called the intermediate value theorem that can help you. The IVT will tell you that if $h$ is continuous (it is) then $h$ will take on all values in between $0$ and $1/4$.
A: *

*$x^2$ has a range of $[0,\infty)$.

*So $e^{x^2}$ has a range of $[1,\infty)$
(knowing that $\exp$ is increasing tells us this).

*So $e^{x^2}+3$ has a range of $[4,\infty)$.

*So $\frac{1}{e^{x^2}+3}$ has a range of $(0,1/4]$
(knowing that the reciprocal function is decreasing and continuous on $[4,\infty)$ tells us this).

A: $e^{x^2}\geq1 $ for $x\geq0$ . So, $e^{x^2} + 3 \geq 4$. If $y \geq4 $, then $1/y \leq 1/4$.
