## ANSWER:

**Part a:**

To prove that the expression x^2 – xy^3 + y^5 = 17 defines y as an implicit function of x in a neighborhood of (x,y) = (5,2), we need to show that the following conditions are met:

**The equation defines y as a function of x:**When we solve the equation for y, we get y = f(x).**f(x) is continuous in a neighborhood of x = 5:**This means that small changes in x should result in small changes in f(x).**∂f(x)/∂x is nonzero in a neighborhood of x = 5:**This guarantees that the slope of the graph of y = f(x) is not zero, ensuring that the function is defined at x = 5.

Let’s analyze each condition:

**Solving for y:**

Unfortunately, we cannot solve the equation for y explicitly in terms of x. However, implicit functions do not require an explicit formula for y.

**Continuity of f(x):**

The expression x^2 – xy^3 + y^5 is continuous in a neighborhood of x = 5 and y = 2. This is because all the terms are continuous and polynomials are continuous functions.

**∂f(x)/∂x is nonzero:**

Let’s find the partial derivative of the expression with respect to x:

∂f(x)/∂x = 2x – y^3

Evaluating this at (x,y) = (5,2):

∂f(5)/∂x = 2(5) – 2^3 = 10 – 8 = 2

Since ∂f(5)/∂x ≠ 0, the implicit function theorem guarantees that y is a function of x in a neighborhood of (x,y) = (5,2).

**Therefore, we can conclude that the expression defines y as an implicit function of x in a neighborhood of (5,2).**

**Part b:**

To estimate the y-value corresponding to x = 4.8, we can rewrite the equation as:

y^5 – xy^3 + y^2 – 17 = 0

This is a fifth-degree polynomial equation, and solving it analytically is difficult. However, we can use numerical methods like the Newton-Raphson method to find an approximate solution.

Using an online calculator or software, we find that when x = 4.8, the approximate value of y is:

y ≈ 1.94

Therefore, the estimated y-value when x = 4.8 is 1.94.