Respuesta :

Answer with explanation:

→A.

[tex]F(x,y)=1 +\sqrt{4-y^2}\\\\F(3,1)=1+\sqrt{4-1^2}\\\\f(3,1)=1+\sqrt{3}[/tex]

→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.

[tex]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[/tex]

Range of F(x, y) is

F(x,y) ≤ -1

And, F(x,y)≥ 3.

Ver imagen Аноним
abidemiokin

The value of f(3, ) is 1 + √3

The domain of the function is y  < ±2 while  the range of the function exists in the range 1≤ f(x)≤3

Domain and range of a function

Given the function expressed as:

[tex]f(x,y)=1+\sqrt{(4-y^2)}[/tex]

To determine the value of f(3, 1)

[tex]f(x,y)=1+\sqrt{(4-1^2)}\\f(x,y)=1+\sqrt{3}\\[/tex]

Hence the value of f(3, ) is 1 + √3

THe domain is the point where the function exists. For the function to exist, then;

4 - y² > 0

4 > y²

y² < 4

y  < ±2

Hence the domain of the function is y  < ±2

Substitute y = 2 into the function

f(2) = 1 + √4-2²
f(2) = 1

Hence the range of the function exists in the range 1≤ f(x)≤3

Learn more on domain and range here: https://brainly.com/question/2264373

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