Question

Let F(x,y)=1+sqrt(4-y^2). . (a) Evaluate F(3,1). (b) Find and sketch the domain of F. (c) Find the range of F

Answer 1

The goal of this function is to get  a positive value under the square root sign so that the value would not be invalid. Evaluating F(3,1), the value is 1 + sqrt(4-1^2) or equal to 1 + sqrt of 3. In this case, the domain would be x equal to any numbers and the range equal to numbers from -2 ≤ y ≤ 2. 

Answer 2

Answer with explanation:

→A.

$F(x,y)=1 +\sqrt{4-y^2}\\\\F(3,1)=1+\sqrt{4-1^2}\\\\f(3,1)=1+\sqrt{3}$

→B.

Domain of F(x,y) will be

→ 4-y²≥0

→2²-y²≥0

→(2-y)(2+y)≥0

→ -2 ≤ y ≤ 2,and x can take any value.

→C.

$F(x,y)-1=\sqrt{4-y^2}\\\\ \text{Squaring both sides}\\\\ (F(x,y)-1)^2=4-y^2\\\\y^2=4 -(F(x,y)-1)^2\\\\y=\sqrt{4 -(F(x,y)-1)^2}\\\\4 -(F(x,y)-1)^2\geq 0\\\\(F(x,y)-1)^2-4\leq 0 \\\\(F(x,y)-1)^2-2^2\leq 0\\\\ {(F(x,y)-1)-2][(F(x,y)-1)+2] \leq 0}\\\\ (F(x,y)-3)[(F(x,y)+1] \leq 0\\\\F(x,y)\leq -1 \text{and} F(x,y)\geq 3$

Range of F(x, y) is

F(x,y) ≤ -1

And, F(x,y)≥ 3.