If `f(x)=-(x|x|)/(1+x^(2))" then "f^(-1)x` equals :

Options

A. `sqrt((|x|)/(1-|x|)`
B. `(sgn (-x)) sqrt((|x|)/(1-|x|)`
C. `-sqrt((x)/(1-x)`
D. `(sgn (x)) sqrt((|x|)/(1+|x|)`

Answer

To find the inverse of the function \( f(x) = -\frac{x |x|}{1 + x^2} \), we will analyze the function for two cases based on the sign of \( x \).

### Step 1: Analyze the function for \( x \geq 0 \)

For \( x \geq 0 \), we have \( |x| = x \). Therefore, the function simplifies to:
\[ f(x) = -\frac{x^2}{1 + x^2} \]
Let \( y = f(x) \):
\[ y = -\frac{x^2}{1 + x^2} \]
Rearranging gives:
\[ y(1 + x^2) = -x^2 \]
\[ y + yx^2 = -x^2 \]
\[ yx^2 + x^2 + y = 0 \]
Factoring out \( x^2 \):
\[ x^2(y + 1) + y = 0 \]
Solving for \( x^2 \):
\[ x^2 = -\frac{y}{y + 1} \]
Taking the square root:
\[ x = \sqrt{-\frac{y}{y + 1}} \]
Thus, for \( x \geq 0 \):
\[ f^{-1}(y) = \sqrt{-\frac{y}{y + 1}} \quad \text{(valid for } y < 0\text{)} \]

### Step 2: Analyze the function for \( x < 0 \)

For \( x < 0 \), we have \( |x| = -x \). Therefore, the function simplifies to:
\[ f(x) = -\frac{-x^2}{1 + x^2} = \frac{x^2}{1 + x^2} \]
Let \( y = f(x) \):
\[ y = \frac{x^2}{1 + x^2} \]
Rearranging gives:
\[ y(1 + x^2) = x^2 \]
\[ y + yx^2 = x^2 \]
\[ yx^2 - x^2 + y = 0 \]
Factoring out \( x^2 \):
\[ x^2(y - 1) + y = 0 \]
Solving for \( x^2 \):
\[ x^2 = -\frac{y}{y - 1} \]
Taking the square root and considering \( x < 0 \):
\[ x = -\sqrt{-\frac{y}{y - 1}} \]
Thus, for \( x < 0 \):
\[ f^{-1}(y) = -\sqrt{-\frac{y}{y - 1}} \quad \text{(valid for } y \geq 0\text{)} \]

### Final Result

Combining both cases, we have:
\[ f^{-1}(y) = \begin{cases} \sqrt{-\frac{y}{y + 1}} & \text{if } y < 0 \\ -\sqrt{-\frac{y}{y - 1}} & \text{if } y \geq 0 \end{cases} \]