Function Composition: Understanding the Basics

This problem requires us to work with function composition, which is a fundamental concept in mathematics. Function composition involves applying one function to the results of another function. We are given an initial function \(f(x)\) and then asked to find new functions \(g(x)\) and \(h(x)\) which are compositions of \(f(x)\) with itself. Finally, we need to calculate the product of these three functions.

The given function is:

• \(f(x) = \frac{1}{1+x}\)

We also have the definitions for \(g(x)\) and \(h(x)\):

• \(g(x) = f\{f(x)\}\)
• \(h(x) = f[f\{f(x)\}]\)

Our goal is to find the value of \(f(x) \cdot g(x) \cdot h(x)\).

Calculating g(x) using Function Composition

To find \(g(x)\), we need to substitute \(f(x)\) into \(f(x)\) itself. This means wherever we see \(x\) in the expression for \(f(x)\), we replace it with the entire expression for \(f(x)\).

We have \(f(x) = \frac{1}{1+x}\).

So, \(g(x) = f\{f(x)\} = f\left(\frac{1}{1+x}\right)\).

Now, substitute \(\frac{1}{1+x}\) for \(x\) in the definition of \(f(x)\):

$$g(x) = \frac{1}{1 + \left(\frac{1}{1+x}\right)}$$

To simplify this expression, we find a common denominator in the denominator:

$$g(x) = \frac{1}{\frac{1(1+x) + 1}{1+x}}$$ $$g(x) = \frac{1}{\frac{1+x+1}{1+x}}$$ $$g(x) = \frac{1}{\frac{x+2}{1+x}}$$

Finally, we invert the denominator and multiply:

$$g(x) = \frac{1+x}{x+2}$$

So, we have successfully calculated \(g(x)\).

Calculating h(x) through Successive Composition

Next, let's find \(h(x)\). The definition states \(h(x) = f[f\{f(x)\}]\). Notice that \(f\{f(x)\}\) is simply \(g(x)\).

Therefore, \(h(x) = f\{g(x)\}\).

We previously found \(g(x) = \frac{1+x}{x+2}\).

Now, substitute \(g(x)\) into \(f(x)\):

$$h(x) = f\left(\frac{1+x}{x+2}\right)$$

Substitute \(\frac{1+x}{x+2}\) for \(x\) in the definition of \(f(x)\):

$$h(x) = \frac{1}{1 + \left(\frac{1+x}{x+2}\right)}$$

Again, find a common denominator in the denominator:

$$h(x) = \frac{1}{\frac{1(x+2) + (1+x)}{x+2}}$$ $$h(x) = \frac{1}{\frac{x+2+1+x}{x+2}}$$ $$h(x) = \frac{1}{\frac{2x+3}{x+2}}$$

Invert the denominator and multiply:

$$h(x) = \frac{x+2}{2x+3}$$

Now we have all three required functions.

Product Calculation: f(x) · g(x) · h(x)

We have the expressions for \(f(x)\), \(g(x)\), and \(h(x)\):

Now, let's find the product \(f(x) \cdot g(x) \cdot h(x)\):

$$f(x) \cdot g(x) \cdot h(x) = \left(\frac{1}{1+x}\right) \cdot \left(\frac{1+x}{x+2}\right) \cdot \left(\frac{x+2}{2x+3}\right)$$

We can observe common terms in the numerator and denominator that will cancel out.

• The term \((1+x)\) in the denominator of \(f(x)\) cancels with the term \((1+x)\) in the numerator of \(g(x)\).
• The term \((x+2)\) in the denominator of \(g(x)\) cancels with the term \((x+2)\) in the numerator of \(h(x)\).

After cancellation, the expression simplifies to:

$$f(x) \cdot g(x) \cdot h(x) = \frac{1}{2x+3}$$

Final Answer Summary

By meticulously applying the rules of function composition and algebraic simplification, we found the product \(f(x) \cdot g(x) \cdot h(x)\).

The calculated value is \(\frac{1}{2x+3}\). This matches one of the provided options.