Composition of Functions

Define the substitution φ(x) = 1/(1−x). Apply it repeatedly:

φ(x) = 1/(1−x)

φ(φ(x)) = 1/(1 − 1/(1−x)) = (x−1)/x

φ(φ(φ(x))) = 1/(1 − (x−1)/x) = x 

Using the original equation at x, φ(x), and φ²(x) respectively:

Eq(i): f(x) + f(1/(1−x)) = 1 + x

Eq(ii): f(1/(1−x)) + f((x−1)/x) = 1 + 1/(1−x)

Eq(iii):f((x−1)/x) + f(x) = 1 + (x−1)/x

Compute (i) − (ii) + (iii), which eliminates f(1/(1−x)) and f((x−1)/x):

2f(x) = (1+x) − (1 + 1/(1−x)) + (1 + (x−1)/x)

2f(x) = 1 + x − 1 − 1/(1−x) + 1 + (x−1)/x

f(x) = (−x³ + x − 1) / [2x(1−x)]

= (x³ − x + 1) / [2x(x−1)]

f(2) = 9/4

f(f(2)) = 829/360

Let's see what happens algebraically when we nest the function inside itself once:

$$f(f(x)) = \frac{f(x)}{\sqrt{1 + 3(f(x))^2}}$$

Substitute $$f(x) = \frac{x}{\sqrt{1 + 3x^2}}$$ into the right side:

$$f(f(x)) = \frac{\frac{x}{\sqrt{1 + 3x^2}}}{\sqrt{1 + 3\left(\frac{x}{\sqrt{1 + 3x^2}}\right)^2}}$$

Simplify the expression inside the square root in the denominator:

$$f(f(x)) = \frac{\frac{x}{\sqrt{1 + 3x^2}}}{\sqrt{1 + \frac{3x^2}{1 + 3x^2}}}$$

Find a common denominator for the terms inside the bottom square root:

$$f(f(x)) = \frac{\frac{x}{\sqrt{1 + 3x^2}}}{\sqrt{\frac{(1 + 3x^2) + 3x^2}{1 + 3x^2}}}$$  $$f(f(x)) = \frac{\frac{x}{\sqrt{1 + 3x^2}}}{\frac{\sqrt{1 + 6x^2}}{\sqrt{1 + 3x^2}}}$$

The $$\sqrt{1 + 3x^2}$$ terms elegantly cancel out:

$$f(f(x)) = \frac{x}{\sqrt{1 + 6x^2}}$$

Notice what happened:

• $$f(x) = \frac{x}{\sqrt{1 + \mathbf{3}x^2}}$$
• $$f(f(x)) = \frac{x}{\sqrt{1 + \mathbf{6}x^2}}$$

Every time we apply the function $$f$$, the coefficient of $$x^2$$ in the denominator simply increases by $$3$$.

Therefore, if we apply the function $$n$$ times, denoted as $$f^{(n)}(x)$$, the logical generalization is:

$$f^{(n)}(x) = \frac{x}{\sqrt{1 + 3nx^2}}$$

The problem defines $$g(x) = f(f(x))$$. Using our pattern, this is $$f^{(2)}(x)$$:

$$g(x) = \frac{x}{\sqrt{1 + 6x^2}}$$

We need to find $$g(g(x))$$. Since $$g(x)$$ applies $$f$$ twice, $$g(g(x))$$ applies $$f$$ a total of four times. Thus, $$g(g(x)) = f^{(4)}(x)$$.

Using our logical pattern formula where $$n = 4$$:

$$g(g(x)) = \frac{x}{\sqrt{1 + 3(4)x^2}} = \frac{x}{\sqrt{1 + 12x^2}}$$

Now we simply substitute $$x = \frac{1}{2}$$ into our generalized formula:

$$g\left(g\left(\frac{1}{2}\right)\right) = \frac{\frac{1}{2}}{\sqrt{1 + 12\left(\frac{1}{2}\right)^2}}$$  $$g\left(g\left(\frac{1}{2}\right)\right) = \frac{\frac{1}{2}}{\sqrt{1 + 12\left(\frac{1}{4}\right)}}$$  $$g\left(g\left(\frac{1}{2}\right)\right) = \frac{\frac{1}{2}}{\sqrt{1 + 3}}$$  $$g\left(g\left(\frac{1}{2}\right)\right) = \frac{\frac{1}{2}}{\sqrt{4}}$$  $$g\left(g\left(\frac{1}{2}\right)\right) = \frac{\frac{1}{2}}{2} = \frac{1}{4}$$

This matches Option C.

Step 1: Understand the Cyclic Nature of the Transformation

The given argument transformation inside the function is $$g(x) = \frac{x-1}{x} = 1 - \frac{1}{x}$$. Let's investigate what happens when we repeatedly compose this transformation:

1. First application: $$g(x) = \frac{x-1}{x}$$
2. Second application: $$g(g(x)) = 1 - \frac{1}{\left(\frac{x-1}{x}\right)} = 1 - \frac{x}{x-1} = \frac{x-1-x}{x-1} = \frac{1}{1-x}$$
3. Third application: $$g(g(g(x))) = 1 - \frac{1}{\left(\frac{1}{1-x}\right)} = 1 - (1-x) = x$$

Because applying the transformation three times returns us exactly back to $$x$$, this function configuration forms a cycle of length 3. This means we can create a closed system of three linear equations to solve for $$f(x)$$ or any specific value directly.

Step 2: Method 1 — Direct Numerical Substitution

Since we specifically need the value at $$x = 2$$, we can plug in numbers directly to see the cyclic sequence manifest:

1. Substitute $$x = 2$$ into the given functional equation:$$f(2) + f\left(\frac{2-1}{2}\right) = 1 + 2$$  $$f(2) + f\left(\frac{1}{2}\right) = 3 \quad \text{--- (Equation 1)}$$
2. Substitute $$x = \frac{1}{2}$$ into the given functional equation:$$f\left(\frac{1}{2}\right) + f\left(\frac{\frac{1}{2}-1}{\frac{1}{2}}\right) = 1 + \frac{1}{2}$$ $$f\left(\frac{1}{2}\right) + f(-1) = \frac{3}{2} \quad \text{--- (Equation 2)}$$
3. Substitute $$x = -1$$ into the given functional equation:$$f(-1) + f\left(\frac{-1-1}{-1}\right) = 1 + (-1)$$ $$f(-1) + f(2) = 0 \quad \text{--- (Equation 3)}$$

Step 3: Solve the Linear System

From Equation 3, we can express $$f(-1)$$ in terms of our target value $$f(2)$$:

$$f(-1) = -f(2)$$

Substitute this back into Equation 2:

$$f\left(\frac{1}{2}\right) - f(2) = \frac{3}{2} \implies f\left(\frac{1}{2}\right) = f(2) + \frac{3}{2}$$

Now, substitute this expression for $$f\left(\frac{1}{2}\right)$$ directly into Equation 1:

$$f(2) + \left(f(2) + \frac{3}{2}\right) = 3$$ $$2f(2) + \frac{3}{2} = 3$$ $$2f(2) = 3 - \frac{3}{2} = \frac{3}{2}$$  $$f(2) = \frac{3}{4}$$

Step 4: Method 2 — General Algebraic Formula (Verification)

Alternatively, we can find the explicit expression for $$f(x)$$ by setting up the general system:

1. $$f(x) + f\left(\frac{x-1}{x}\right) = 1 + x$$
2. $$f\left(\frac{x-1}{x}\right) + f\left(\frac{1}{1-x}\right) = 1 + \frac{x-1}{x} = 2 - \frac{1}{x}$$
3. $$f\left(\frac{1}{1-x}\right) + f(x) = 1 + \frac{1}{1-x}$$

Performing the linear combination $$\text{(Eq 1)} + \text{(Eq 3)} - \text{(Eq 2)}$$ cancels out the cyclic terms, leaving:

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

Plugging in $$x = 2$$:

$$2f(2) = 2 + \frac{1}{2} + \frac{1}{1-2} = 2 + \frac{1}{2} - 1 = \frac{3}{2} \implies f(2) = \frac{3}{4}$$

Step 5: Compute the Final Requested Expression

The question asks for the value of $$4f(2)$$:

$$4f(2) = 4 \times \frac{3}{4} = 3$$