If `f:R to R ` be defined by `f(x) = (x)/(sqrt(1 +x^(2) )),` then `(fofof)(x)= …………. .`

Explanation

To solve the problem, we need to find \( (f \circ f \circ f)(x) \) where \( f(x) = \frac{x}{\sqrt{1 + x^2}} \).

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

1. Start with the function:
\[ f(x) = \frac{x}{\sqrt{1 + x^2}} \]

2. Now, substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{x}{\sqrt{1 + x^2}}\right) \]

3. Replace \( x \) in \( f(x) \) with \( \frac{x}{\sqrt{1 + x^2}} \):
\[ f(f(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2}} \]

4. Simplifying the denominator:
\[ \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = \frac{x^2}{1 + x^2} \]
Therefore,
\[ 1 + \left(\frac{x}{\sqrt{1 + x^2}}\right)^2 = 1 + \frac{x^2}{1 + x^2} = \frac{1 + x^2 + x^2}{1 + x^2} = \frac{1 + 2x^2}{1 + x^2} \]

5. So, we have:
\[ f(f(x)) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{\frac{1 + 2x^2}{1 + x^2}}} = \frac{x}{\sqrt{1 + x^2}} \cdot \frac{\sqrt{1 + x^2}}{\sqrt{1 + 2x^2}} = \frac{x}{\sqrt{1 + 2x^2}} \]

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

1. Now, we need to find \( f(f(f(x))) \):
\[ f(f(f(x))) = f\left(f(f(x))\right) = f\left(\frac{x}{\sqrt{1 + 2x^2}}\right) \]

2. Substitute \( \frac{x}{\sqrt{1 + 2x^2}} \) into \( f(x) \):
\[ f(f(f(x))) = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{1 + \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2}} \]

3. Simplifying the denominator:
\[ \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2 = \frac{x^2}{1 + 2x^2} \]
Therefore,
\[ 1 + \left(\frac{x}{\sqrt{1 + 2x^2}}\right)^2 = 1 + \frac{x^2}{1 + 2x^2} = \frac{1 + 2x^2 + x^2}{1 + 2x^2} = \frac{1 + 3x^2}{1 + 2x^2} \]

4. So, we have:
\[ f(f(f(x))) = \frac{\frac{x}{\sqrt{1 + 2x^2}}}{\sqrt{\frac{1 + 3x^2}{1 + 2x^2}}} = \frac{x}{\sqrt{1 + 2x^2}} \cdot \frac{\sqrt{1 + 2x^2}}{\sqrt{1 + 3x^2}} = \frac{x}{\sqrt{1 + 3x^2}} \]

### Final Answer
Thus, the final result is:
\[ (f \circ f \circ f)(x) = \frac{x}{\sqrt{1 + 3x^2}} \]