Step-by-step Solution:

We are given the function: \[ f(x) = (x - 1)^2 (x + 1)^3 \] \[\] Step 1: Take the derivative of \(f(x)\) \[\] To find the maximum value of \(f(x)\), we first find the critical points by computing the first derivative of \(f(x)\) and setting it equal to zero. \[\] Using the product rule: \[ f'(x) = \frac{d}{dx} \left( (x - 1)^2 (x + 1)^3 \right) \] Let \(u = (x - 1)^2\) and \(v = (x + 1)^3\). The product rule states that: \[ f'(x) = u'v + uv' \] First, compute the derivatives of \(u\) and \(v\): \[ u' = 2(x - 1), \quad v' = 3(x + 1)^2 \] Now, substitute these into the product rule: \[ f'(x) = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] Factor out the common term \((x - 1)(x + 1)^2\): \[ f'(x) = (x - 1)(x + 1)^2 \left( 2(x + 1) + 3(x - 1) \right) \] Simplify the terms inside the parentheses: \[ 2(x + 1) + 3(x - 1) = 2x + 2 + 3x - 3 = 5x - 1 \] Thus, the derivative is: \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \] Step 2:To find the critical points, set \(f'(x) = 0\): \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \] This gives three factors to solve for: \[\] 1. \(x - 1 = 0 \Rightarrow x = 1\) \[\] 2. \((x + 1)^2 = 0 \Rightarrow x = -1\) \[\] 3. \(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\) \[\] Step 3:Now, evaluate \(f(x)\) at the critical points \(x = 1\), \(x = -1\), and \(x = \frac{1}{5}\): \[\] 1. At \(x = 1\): \[ f(1) = (1 - 1)^2 (1 + 1)^3 = 0 \cdot 8 = 0 \] 2. At \(x = -1\): \[ f(-1) = (-1 - 1)^2 (-1 + 1)^3 = 4 \cdot 0 = 0 \] 3. At \(x = \frac{1}{5}\): \[ f\left( \frac{1}{5} \right) = \left( \frac{1}{5} - 1 \right)^2 \left( \frac{1}{5} + 1 \right)^3 = \left( \frac{-4}{5} \right)^2 \left( \frac{6}{5} \right)^3 \] Simplify each term: \[ \left( \frac{-4}{5} \right)^2 = \frac{16}{25}, \quad \left( \frac{6}{5} \right)^3 = \frac{216}{125} \] Now multiply these: \[ f\left( \frac{1}{5} \right) = \frac{16}{25} \times \frac{216}{125} = \frac{16 \times 216}{25 \times 125} = \frac{3456}{3125} \] Thus, the maximum value of \(f(x)\) is: \[ f\left( \frac{1}{5} \right) = \frac{3456}{3125} \] Step 4: Express in the form \(\frac{2^p 3^q}{3125}\) \[\] Now, express \(3456\) in terms of powers of 2 and 3: \[ 3456 = 2^7 \times 3^3 \] Thus: \[ f\left( \frac{1}{5} \right) = \frac{2^7 \times 3^3}{3125} \] We are given that the maximum value is in the form \(\frac{2^p 3^q}{3125}\). Comparing this with the above expression, we find that: \[ p = 7, \quad q = 3 \] Final Answer: The ordered pair \((p, q)\) is: \[ {(7, 3)} \]