# How do you find the domain and range of h(x) = e^(-x^2)?

The domain is the set of all real number hence ${D}_{f} = \left(- \infty , + \infty\right)$

The range is ${R}_{f} = \left(0 , 1\right]$

Because ${e}^{{x}^{2}} \ge {x}^{2} + 1$ hence $\frac{1}{e} ^ \left({x}^{2}\right) \le \frac{1}{{x}^{2} + 1} \le 1$

but always $\frac{1}{e} ^ \left({x}^{2}\right) > 0$