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 step by step, we will first define the functions and then compute their values accordingly.

### Step 1: Define the function \( f(x) \)
We are given:
\[ f(x) = \frac{1}{1 - x} \]

### Step 2: Calculate \( g(x) = f(f(x)) \)
To find \( g(x) \), we need to substitute \( f(x) \) into itself:
\[ g(x) = f(f(x)) = f\left(\frac{1}{1 - x}\right) \]
Now, we substitute \( \frac{1}{1 - x} \) into \( f(x) \):
\[ g(x) = \frac{1}{1 - \frac{1}{1 - x}} \]
To simplify this, we first find the denominator:
\[ 1 - \frac{1}{1 - x} = \frac{(1 - x) - 1}{1 - x} = \frac{-x}{1 - x} \]
Thus,
\[ g(x) = \frac{1}{\frac{-x}{1 - x}} = \frac{1 - x}{-x} = \frac{x - 1}{x} \]

### Step 3: Calculate \( h(x) = f(g(x)) \)
Now we need to find \( h(x) \):
\[ h(x) = f(g(x)) = f\left(\frac{x - 1}{x}\right) \]
Substituting \( \frac{x - 1}{x} \) into \( f(x) \):
\[ h(x) = \frac{1}{1 - \frac{x - 1}{x}} = \frac{1}{\frac{x - (x - 1)}{x}} = \frac{1}{\frac{1}{x}} = x \]

### Step 4: Calculate the product \( f(x) \cdot g(x) \cdot h(x) \)
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:
\[ = \frac{1}{1 - x} \cdot \frac{x - 1}{x} \cdot x \]
The \( x \) in the numerator and denominator cancels out:
\[ = \frac{1}{1 - x} \cdot (x - 1) = \frac{-(1 - x)}{1 - x} = -1 \]

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