Function Condition Analysis

The problem asks about the properties of a real-valued function \( g(x) \) defined on an interval containing 0, such that \( f(x, g(x)) = 0 \), where \( f(x, y) = x^2 - y^3 \). The condition \( f(x, g(x)) = 0 \) translates to:

$$ x^2 - [g(x)]^3 = 0 $$

Rearranging this equation to solve for \( g(x) \), we get:

$$ [g(x)]^3 = x^2 $$

Taking the cube root of both sides gives us the unique real-valued solution for \( g(x) \):

$$ g(x) = \sqrt[3]{x^2} $$

This function can also be written as \( g(x) = x^{2/3} \).

Now we need to analyze the continuity and differentiability of this function \( g(x) = x^{2/3} \) on an interval that includes 0.

Continuity of g(x) at 0

A function is continuous at a point if the limit of the function as it approaches the point exists and is equal to the function's value at that point. For \( g(x) = x^{2/3} \) at \( x=0 \):

• The function is defined at \( x=0 \), and \( g(0) = 0^{2/3} = 0 \).
• The limit as \( x \) approaches 0 is \( \lim_{x \to 0} x^{2/3} \). As \( x \) gets closer to 0, \( x^{2/3} \) gets closer to 0. So, \( \lim_{x \to 0} x^{2/3} = 0 \).

Since \( \lim_{x \to 0} g(x) = g(0) \), the function \( g(x) = x^{2/3} \) is continuous at \( x=0 \). In fact, the function \( g(x) = x^{2/3} \) is continuous for all real numbers.

Since for each \( x \), there is only one real value \( y = \sqrt[3]{x^2} \) satisfying \( y^3 = x^2 \), the function \( g(x) = x^{2/3} \) is the only real-valued function satisfying the condition \( f(x, g(x)) = 0 \). Because this function is continuous on any interval containing 0, there is exactly one such continuous function.

Differentiability of g(x) at 0

A function \( g(x) \) is differentiable at \( x=0 \) if the limit of the difference quotient exists as \( h \to 0 \):

$$ g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} $$

For \( g(x) = x^{2/3} \):

$$ g'(0) = \lim_{h \to 0} \frac{h^{2/3} - 0^{2/3}}{h} = \lim_{h \to 0} \frac{h^{2/3}}{h} = \lim_{h \to 0} h^{2/3 - 1} = \lim_{h \to 0} h^{-1/3} = \lim_{h \to 0} \frac{1}{h^{1/3}} $$

This limit does not exist because as \( h \to 0^+ \), \( 1/h^{1/3} \to +\infty \), and as \( h \to 0^- \), \( 1/h^{1/3} \to -\infty \). Therefore, the function \( g(x) = x^{2/3} \) is not differentiable at \( x=0 \).

Since \( g(x) = x^{2/3} \) is the unique real-valued function satisfying \( f(x, g(x)) = 0 \), and it is not differentiable at \( x=0 \), there is no differentiable real-valued function \( g \) defined on an interval containing 0 such that \( f(x, g(x)) = 0 \).

Evaluating the Statements

Let's examine the given statements based on our analysis of \( g(x) = x^{2/3} \):

1. There is no continuous real-valued function g defined on any interval of \( \reals \) containing 0 such that f(x, g(x)) = 0.
   This is false, because \( g(x) = x^{2/3} \) is continuous and satisfies the condition.
2. There is exactly one continuous real-valued function g defined on an interval of \( \reals \) containing 0 such that f(x, g(x)) = 0.
   This is true. We found the unique function \( g(x) = x^{2/3} \) which is continuous on any interval containing 0.
3. There is exactly one differentiable real-valued function g defined on an interval of \( \reals \) containing 0 such that f(x, g(x)) = 0.
   This is false. The unique function \( g(x) = x^{2/3} \) is not differentiable at 0.
4. There are two distinct differentiable real-valued functions g on an interval of \( \reals \) containing 0 such that f(x, g(x)) = 0.
   This is false. There is only one real-valued function satisfying the condition, and it is not differentiable at 0.

Based on the analysis, only the second statement is true.