Consider the function defined implicitly by the equation `y^3-3y+x=0` on various intervals in the real line. If `x in (-oo,-2) uu (2,oo)`, the equation implicitly defines a unique real-valued defferentiable function `y=f(x)`. If `x in (-2,2)`, the equation implicitly defines a unique real-valud diferentiable function `y-g(x)` satisfying `g_(0)=0`.
If `f(-10sqrt2)=2sqrt(2)`, then `f"(-10sqrt(2))` is equal to

Options

A. (a)`(4sqrt(2))/(7^(3)3^(2))`
B. (b)`-(4sqrt(2))/(7^(3)3^(2))`
C. (c)`(4sqrt(2))/(7^(3)3)`
D. (d)`-(4sqrt(2))/(7^(3)3)`

Answer

To solve the problem, we begin with the implicit function defined by the equation:

\[ y^3 - 3y + x = 0 \]

We need to find the second derivative \( f''(-10\sqrt{2}) \) given that \( f(-10\sqrt{2}) = 2\sqrt{2} \).

### Step 1: Differentiate the equation implicitly
We differentiate both sides of the equation with respect to \( x \):

\[ \frac{d}{dx}(y^3) - \frac{d}{dx}(3y) + \frac{d}{dx}(x) = 0 \]

Using the chain rule, we get:

\[ 3y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} + 1 = 0 \]

### Step 2: Solve for \( \frac{dy}{dx} \)
Rearranging the equation gives:

\[ (3y^2 - 3) \frac{dy}{dx} + 1 = 0 \]

Thus,

\[ \frac{dy}{dx} = -\frac{1}{3y^2 - 3} \]

### Step 3: Evaluate \( \frac{dy}{dx} \) at \( x = -10\sqrt{2} \)
We know \( f(-10\sqrt{2}) = 2\sqrt{2} \). We substitute \( y = 2\sqrt{2} \) into the derivative:

\[ \frac{dy}{dx} = -\frac{1}{3(2\sqrt{2})^2 - 3} \]

Calculating \( (2\sqrt{2})^2 \):

\[ (2\sqrt{2})^2 = 8 \quad \Rightarrow \quad 3(2\sqrt{2})^2 - 3 = 3(8) - 3 = 24 - 3 = 21 \]

So,

\[ \frac{dy}{dx} = -\frac{1}{21} \]

### Step 4: Differentiate again to find \( \frac{d^2y}{dx^2} \)
We differentiate \( \frac{dy}{dx} \) again with respect to \( x \):

Using the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{1}{3y^2 - 3}\right) = \frac{(0)(3y^2 - 3) - (-1)(6y \frac{dy}{dx})}{(3y^2 - 3)^2} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = \frac{6y \frac{dy}{dx}}{(3y^2 - 3)^2} \]

### Step 5: Substitute \( y = 2\sqrt{2} \) and \( \frac{dy}{dx} = -\frac{1}{21} \)
Now substituting \( y = 2\sqrt{2} \) and \( \frac{dy}{dx} = -\frac{1}{21} \):

\[ \frac{d^2y}{dx^2} = \frac{6(2\sqrt{2})(-\frac{1}{21})}{(21)^2} \]

Calculating \( 6(2\sqrt{2})(-\frac{1}{21}) \):

\[ = -\frac{12\sqrt{2}}{21} \]

Now, substituting into the denominator:

\[ (21)^2 = 441 \]

Thus,

\[ \frac{d^2y}{dx^2} = \frac{-12\sqrt{2}}{441} = -\frac{4\sqrt{2}}{147} \]

### Final Answer
Therefore, the value of \( f''(-10\sqrt{2}) \) is:

\[ f''(-10\sqrt{2}) = -\frac{4\sqrt{2}}{147} \]