`g(x+y)=g(x)+g(y)+3xy(x+y)AA x, y in R" and "g'(0)=-4.`
For which of the following values of x is `sqrt(g(x))` not defined ?

Options

A. `[-2,0]`
B. `[2,oo)`
C. `[-1,1]`
D. none of these

Explanation

To solve the problem, we need to analyze the function \( g(x) \) defined by the equation:

\[ g(x+y) = g(x) + g(y) + 3xy(x+y) \]

We also know that \( g'(0) = -4 \).

### Step 1: Differentiate the given equation

We start by differentiating both sides of the equation with respect to \( x \):

\[ \frac{d}{dx}[g(x+y)] = \frac{d}{dx}[g(x) + g(y) + 3xy(x+y)] \]

Using the chain rule on the left side and the product rule on the right side, we get:

\[ g'(x+y) = g'(x) + 3y(x+y) + 3xy \]

### Step 2: Substitute \( x = 0 \)

Now, we substitute \( x = 0 \):

\[ g'(y) = g'(0) + 3y(0+y) + 0 \]

Given \( g'(0) = -4 \), we have:

\[ g'(y) = -4 + 3y^2 \]

### Step 3: Integrate to find \( g(y) \)

Next, we integrate \( g'(y) \) to find \( g(y) \):

\[ g(y) = \int (-4 + 3y^2) \, dy = -4y + y^3 + C \]

Where \( C \) is a constant.

### Step 4: Find \( g(0) \)

To find the constant \( C \), we can use the original equation by setting \( x = 0 \) and \( y = 0 \):

\[ g(0) = g(0) + g(0) + 0 \]

This implies \( g(0) = 0 \). Therefore, we set \( g(0) = -4(0) + 0^3 + C = 0 \), which gives \( C = 0 \).

### Step 5: Write the final form of \( g(x) \)

Thus, we have:

\[ g(x) = -4x + x^3 \]

### Step 6: Find when \( \sqrt{g(x)} \) is not defined

The expression \( \sqrt{g(x)} \) is not defined when \( g(x) < 0 \). We need to find the values of \( x \) for which:

\[ -4x + x^3 < 0 \]

This can be rearranged to:

\[ x^3 - 4x < 0 \]

Factoring gives:

\[ x(x^2 - 4) < 0 \quad \Rightarrow \quad x(x-2)(x+2) < 0 \]

### Step 7: Analyze the intervals

To find the intervals where this inequality holds, we can analyze the sign changes around the roots \( x = -2, 0, 2 \):

- For \( x < -2 \): All factors are negative, so the product is negative.
- For \( -2 < x < 0 \): \( x \) is negative, \( (x-2) \) is negative, \( (x+2) \) is positive, so the product is positive.
- For \( 0 < x < 2 \): \( x \) is positive, \( (x-2) \) is negative, \( (x+2) \) is positive, so the product is negative.
- For \( x > 2 \): All factors are positive, so the product is positive.

Thus, \( g(x) < 0 \) in the intervals \( (-\infty, -2) \) and \( (0, 2) \).

### Conclusion

The values of \( x \) for which \( \sqrt{g(x)} \) is not defined are:

- \( x \in (-\infty, -2) \)
- \( x \in (0, 2) \)