Question

What is the value of the product $f(2) \cdot f^2(2) \cdot f^3(2) \cdot f^4(2) \cdot f^5(2)$ ?

• A. $1/3$
• B. $3$
• C. $1/18$
• D. None of these
• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The Correct Option is C

Given: \[ f(x) = \begin{cases} \frac{1}{1+x}, & x \ge 0
1 + x, & x<0 \end{cases} \] Also $f^n(x) = f(f^{n-1}(x))$. Step 1: $f(2) = \frac{1}{1+2} = \frac{1}{3}$. Step 2: $f^2(2) = f\left(\frac{1}{3}\right) = \frac{1}{1+\frac{1}{3}} = \frac{3}{4}$. Step 3: $f^3(2) = f\left(\frac{3}{4}\right) = \frac{1}{1+\frac{3}{4}} = \frac{4}{7}$. Step 4: $f^4(2) = f\left(\frac{4}{7}\right) = \frac{1}{1+\frac{4}{7}} = \frac{7}{11}$. Step 5: $f^5(2) = f\left(\frac{7}{11}\right) = \frac{1}{1+\frac{7}{11}} = \frac{11}{18}$. Product: \[ \frac{1}{3} \cdot \frac{3}{4} \cdot \frac{4}{7} \cdot \frac{7}{11} \cdot \frac{11}{18} = \frac{1}{18}. \] \[ \boxed{\frac{1}{18}} \]

Answer 2

We are given:

• \( r \geq 2 \)
• Starting value: \( x_0 = -r < 0 \)

Step 1: First iteration

Applying the function once: \[ f(x_0) = 1 + x_0 = 1 - r \] Since \( r \geq 2 \), we have \( 1 - r \leq -1 < 0 \). Therefore, after the first step, the value remains negative.

Step 2: General formula for negative stage

While the value remains negative, the iteration rule is: \[ f(x) = 1 + x \] Therefore, for \( k \leq r \): \[ f^k(-r) = -r + k \]

Step 3: Reaching zero

When \( k = r - 1 \): \[ f^{r-1}(-r) = -r + (r - 1) = -1 \] When \( k = r \): \[ f^{r}(-r) = -r + r = 0 \]

Step 4: Transition to positive rule

For non-negative inputs, we use: \[ f(x) = \frac{1}{1 + x} \] Therefore: \[ f^{r+1}(-r) = f(0) = \frac{1}{1 + 0} = 1 \]

Step 5: Summation

The sum of the three consecutive values is: \[ (-1) + 0 + 1 = 0 \] This confirms that: \[ \boxed{0} \]

Step 6: Final conclusion

The sum is always \(0\) for \( r \geq 2 \), matching the correct option (2).