Answer
a) $e$
b) $D=\{(x,y)\in \mathbb{R}^2|y\ge x^2\}$
c) $R=\{z\in \mathbb{R}|z\ge1\}$
Work Step by Step
a) $h(-2, 5) = e^{\sqrt{y-x^2}}= e^{\sqrt{5-4}}= e^{\sqrt{1}}=e$
b) The value under the square root must be greater than or equal to 0. Thus, $y-x^2\ge0$, which includes all points above or at the parabola $y=x^2$.
c) The smallest value of $\sqrt{y-x^2}$ is 0, which means the smallest value of the function h is 1. The value inside the square root can increase to infinity, which means the range of the function h does not have an upper bound.
