# Chapter 14 - Partial Derivatives - 14.1 Functions of Several Variables - 14.1 Exercises - Page 940: 10

(a) $F(3,1)=1+\sqrt{3}$. (b) Domain: $\mathbb{R}×[-2,2]$. It is on the figure below (blue region, infinite in $x$ direction). (c) Range: $[1,3]$. (a) By direct substitution $x\to3$ and $y\to1$ we have $$F(3,1)=1+\sqrt{4-1^2}=1+\sqrt{3}.$$ (b) The argument of the square root must be non negative so we demand that $4-y^2\geq0$ giving $y^2\leq4$ and this gives $|y|\leq2$, so $y\in[-2,2]$. This means that the domain is an infinite (in $x$ direction) rectangle, including its edges, with $-2\leq y\leq2$ or, in other words, $\mathbb{R}×[-2,2].$ The sketch in on the figure below. (c) For the range, notice that when $|y|$ increases then $1+\sqrt{4-y^2}$ decreases. So the maximal $|y|$, which is $2$ gives the smallest value of $F$ and conversely, the smallest value of $|y|$ which is $0$ gives the maximal value of $F$. Thus, we have $F_{min}=1+\sqrt{4-2^2}=1$ and $F_{max}=1+\sqrt{4+0^2}=3.$ Between these bounds, $F$ changes continuously so we have for the range $[1,3]$. 