If `f(x)=(x)/(sqrt(1+x^(2)))`, then `fofof(x)` is equal to:

Options

A. `(x)/(sqrt((1+3x^(2))))`q
B. `(3x)/(sqrt(1-x^(2)))`
C. `(3x)/(sqrt((1+x^(2)))`
D. none of these

Answer

To find \( f(f(f(x))) \) where \( f(x) = \frac{x}{\sqrt{1+x^2}} \), we will follow these steps:

### Step 1: Find \( f(f(x)) \)

1. Start with the function \( f(x) = \frac{x}{\sqrt{1+x^2}} \).
2. Substitute \( f(x) \) into itself to find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{x}{\sqrt{1+x^2}}\right) \]
3. Now, replace \( x \) in the function \( f \) with \( \frac{x}{\sqrt{1+x^2}} \):
\[ f\left(\frac{x}{\sqrt{1+x^2}}\right) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+x^2}}\right)^2}} \]
4. Simplify the expression inside the square root:
\[ 1+\left(\frac{x}{\sqrt{1+x^2}}\right)^2 = 1+\frac{x^2}{1+x^2} = \frac{1+x^2+x^2}{1+x^2} = \frac{1+2x^2}{1+x^2} \]
5. Therefore, we have:
\[ f(f(x)) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{\frac{1+2x^2}{1+x^2}}} = \frac{x}{\sqrt{1+x^2}} \cdot \sqrt{\frac{1+x^2}{1+2x^2}} = \frac{x}{\sqrt{1+2x^2}} \]

### Step 2: Find \( f(f(f(x))) \)

1. Now we need to find \( f(f(f(x))) = f\left(f(f(x))\right) = f\left(\frac{x}{\sqrt{1+2x^2}}\right) \).
2. Substitute \( \frac{x}{\sqrt{1+2x^2}} \) into \( f \):
\[ f\left(\frac{x}{\sqrt{1+2x^2}}\right) = \frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+2x^2}}\right)^2}} \]
3. Simplify the expression inside the square root:
\[ 1+\left(\frac{x}{\sqrt{1+2x^2}}\right)^2 = 1+\frac{x^2}{1+2x^2} = \frac{1+2x^2+x^2}{1+2x^2} = \frac{1+3x^2}{1+2x^2} \]
4. Therefore, we have:
\[ f(f(f(x))) = \frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{\frac{1+3x^2}{1+2x^2}}} = \frac{x}{\sqrt{1+2x^2}} \cdot \sqrt{\frac{1+2x^2}{1+3x^2}} = \frac{x}{\sqrt{1+3x^2}} \]

### Final Answer

Thus, we conclude that:
\[ f(f(f(x))) = \frac{x}{\sqrt{1+3x^2}} \]