 Solution

The derivative is $f^{\prime }(x)=-4x^3{e}^{-x^4}$. Its absolute value is continuous and tends to zero as $x\to \pm \infty$, so it is bounded. More explicitly,obtained by differentiating $4t^3{e}^{-t^4}$ for $t\ge 0$. The mean value theorem gives $|f(x)-f(y)|\le M|x-y|$ for every $x,y$. Thus $f$ is Lipschitz continuous, and choosing $\delta =\epsilon /M$ proves $e^{-x^4}$ is uniformly continuous on $\mathbb{R}$.