$f(x)=e^{x^2} +2x(xe^{x^2}-e^{-x^2})$ and $x\in[0,1]$
I want to know how $f(x)$ is positive in the interval.
$e^{x^2}$ is +ve but $(xe^{x^2}-e^{-x^2})$ is not always +ve.
So how can I show that $e^{x^2}$ is always greater than $2x(xe^{x^2}-e^{-x^2})$
Thanks.
It is enough to show that $\frac {f(x)} {e^{x^{2}}} > 0$. This reduces to $1+2x^{2}-2xe^{-2x^{2}} >0$. Re-write this as $e^{2x^{2}} >\frac {2x} {1+2x^{2}}$. To prove this use the following:
a) $e^{2x^{2}}>1+2x^{2}$ (from the series expansion)
b) $(2x-1)^{2} \geq 0$.
