Question

Let f(x y)= 1 + sqrt(4 - y²)

-evaluate f(3, 1)
-find and sketch the domain of f
-find the range of f

Answer 1

Evaluate f(3, 1) by substituting 1 for y, resulting in 1 + sqrt(3). The function's domain is all x and y where -2 ≤ y ≤ 2, and the range is from 1 to 3.

Step-by-step explanation:

You asked to evaluate the function f(x, y) = 1 + sqrt(4 - y²) at the point (3, 1), to find and sketch the domain of f, and to find the range of f.

1. To evaluate f(3, 1), we substitute y with 1 into the function:
2. f(3, 1) = 1 + sqrt(4 - 1²) = 1 + sqrt(4 - 1) = 1 + sqrt(3).
3. The domain of f is all the (x, y) pairs where the expression inside the square root, 4 - y², is non-negative. This means y² ≤ 4, so -2 ≤ y ≤ 2. X can take any real value since it does not appear in the function.
4. The range of f is the set of possible output values. Since the square root function outputs values from 0 to 2 for the domain of y mentioned, and we add 1 to this, the range of f is from 1 to 3.