If `f(x)=(1)/((1-x)),g(x)=f{f(x)}andh(x)=f[f{f(x)}]`. Then the value of f(x).g(x).h(x) is

Options

A. 6
B. -1
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( f(x) \cdot g(x) \cdot h(x) \) where:

1. \( f(x) = \frac{1}{1-x} \)
2. \( g(x) = f(f(x)) \)
3. \( h(x) = f(f(f(x))) \)

Let's go through the calculations step by step.

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

Given:
\[ f(x) = \frac{1}{1-x} \]

### Step 2: Calculate \( g(x) = f(f(x)) \)

First, we need to find \( f(f(x)) \):
\[ f(f(x)) = f\left(\frac{1}{1-x}\right) \]

Now, substitute \( \frac{1}{1-x} \) into \( f(x) \):
\[ f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} \]

Simplifying the denominator:
\[ 1 - \frac{1}{1-x} = \frac{(1-x) - 1}{1-x} = \frac{-x}{1-x} \]

Thus,
\[ f(f(x)) = \frac{1}{\frac{-x}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x} \]

So, we have:
\[ g(x) = \frac{x-1}{x} \]

### Step 3: Calculate \( h(x) = f(f(f(x))) \)

Now, we need to find \( f(f(f(x))) = f(g(x)) = f\left(\frac{x-1}{x}\right) \):
\[ f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} \]

Simplifying the denominator:
\[ 1 - \frac{x-1}{x} = \frac{x - (x-1)}{x} = \frac{1}{x} \]

Thus,
\[ f(f(f(x))) = \frac{1}{\frac{1}{x}} = x \]

So, we have:
\[ h(x) = x \]

### Step 4: Calculate \( f(x) \cdot g(x) \cdot h(x) \)

Now we can find the product:
\[ f(x) \cdot g(x) \cdot h(x) = \left(\frac{1}{1-x}\right) \cdot \left(\frac{x-1}{x}\right) \cdot x \]

Simplifying this:
\[ = \frac{1}{1-x} \cdot \frac{x-1}{x} \cdot x = \frac{(x-1)}{(1-x)} = -1 \]

### Final Answer

Thus, the value of \( f(x) \cdot g(x) \cdot h(x) \) is:
\[ \boxed{-1} \]