Step-by-step Solution:

Step 1: Find the Greatest Value of \( f(x) \) in \([-2,3]\) \[\] We first find the critical points by differentiating \( f(x) \): \[ f'(x) = \frac{d}{dx} (x^3 + 3x - 9) = 3x^2 + 3 \] Setting \( f'(x) = 0 \): \[ 3x^2 + 3 = 0 \] \[ x^2 = -1 \] Since this has no real solution, there are no critical points in the given interval. \[\] Therefore, the maximum value occurs at one of the endpoints \( x = -2 \) or \( x = 3 \). \[\] Evaluate \( f(x) \) at the endpoints: \[ f(-2) = (-2)^3 + 3(-2) - 9 = -8 - 6 - 9 = -23 \] \[ f(3) = (3)^3 + 3(3) - 9 = 27 + 9 - 9 = 27 \] The greatest value in the interval is 27. \[\] Step 2: Find \( f'(0) \) \[\] We substitute \( x = 0 \) into \( f'(x) \): \[ f'(0) = 3(0)^2 + 3 = 3 \] Step 3: Use the Sum Formula of an Infinite GP \[\] The sum of an infinite decreasing GP is given by: \[ S = \frac{a}{1 - r} \] We are given that: \[ S = 27 \] The first term is \( a \), and the difference between the first two terms is: \[ a - ar = 3 \] Step 4: Solve for \( r \) \[\] Using the sum equation: \[ \frac{a}{1 - r} = 27 \] \[ a = 27(1 - r) \] Using the second equation: \[ a(1 - r) = 3 \] Substituting \( a = 27(1 - r) \): \[ 27(1 - r)(1 - r) = 3 \] \[ 27(1 - 2r + r^2) = 3 \] \[ 27 - 54r + 27r^2 = 3 \] \[ 27r^2 - 54r + 24 = 0 \] Dividing by 3: \[ 9r^2 - 18r + 8 = 0 \] Step 5: Solve the Quadratic Equation \[\] Using the quadratic formula: \[ r = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(9)(8)}}{2(9)} \] \[ r = \frac{18 \pm \sqrt{324 - 288}}{18} \] \[ r = \frac{18 \pm \sqrt{36}}{18} \] \[ r = \frac{18 \pm 6}{18} \] \[ r = \frac{18 + 6}{18} = \frac{24}{18} = \frac{4}{3} \quad \text{(not valid, as \( |r| < 1 \) for decreasing GP)} \] \[ r = \frac{18 - 6}{18} = \frac{12}{18} = \frac{2}{3} \] Thus, the common ratio is: \[ \frac{2}{3} \]