Question

Consider F: R3 → R2 defined by F(x, y, z) = (2+1)2 + y -1, xy + y). Using the Implicit Function Theorem, we can conclude that set S = {(x, y, z) € R3: F(x, y, z) = (0,0)} can be locally described by the graph of Cl functions in some neighbourhood of (0,0,0). By the theorem, we are guaranteed the existence of a suitable C1 function of which form? [2 1 0 Hint: DF (0,0,0) 0 1 0 [ 1 ]

a. x = 4(y, z)
b. y = 4(x,z)
c. z=(x, y)
d. (x, y) = 4(2)
e. (x,z) = 4(y)
f. (y, z) = 4(x)

Answer 1

The set S can be locally described by the graph of a C^1 function in the neighborhood of (0,0,0) using the Implicit Function Theorem.

Step-by-step explanation:

The Implicit Function Theorem states that if a function F: R^3 -> R^2 is continuously differentiable and F(a,b,c) = (0,0), where (a,b,c) is some point in R^3, then in a neighborhood of (a,b,c), the set S = {(x,y,z) in R^3: F(x,y,z) = (0,0)} can be locally described by the graph of C^1 functions.

In this case, we are given F(x,y,z) = (2x+1)^2 + y - 1, (xy + y).

We can use the Implicit Function Theorem to conclude that a suitable C^1 function that describes the set S in the neighborhood of (0,0,0) has the form y = 4x(2x^2-1).