Step-by-step Solution:

Let's solve the problem step by step.
Step 1: Find the Greatest Value of \( f(x) \) in the Interval \( [-2,3] \)
The function given is: \[ f(x) = x^3 + 3x - 9 \] To find its greatest value in the interval \([-2,3]\), 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 \( x^2 = -1 \) has no real solution, there are no critical points. So, we only check \( f(x) \) at the endpoints \( x = -2 \) and \( x = 3 \).
Evaluate at \( x = -2 \): \[ f(-2) = (-2)^3 + 3(-2) - 9 \] \[ = -8 - 6 - 9 = -23 \] Evaluate at \( x = 3 \): \[ f(3) = (3)^3 + 3(3) - 9 \] \[ = 27 + 9 - 9 = 27 \] Thus, the greatest value of \( f(x) \) in the given interval is 27.
Step 2: Find \( f'(0) \)
We already found \( f'(x) = 3x^2 + 3 \). Substituting \( x = 0 \): \[ f'(0) = 3(0)^2 + 3 = 3 \] Step 3: Use the Sum of an Infinite GP Formula
For an infinite decreasing GP, the sum is given by: \[ S_{\infty} = \frac{a}{1 - r} \] From the problem statement: \[ S_{\infty} = 27, \quad a_1 - a_2 = 3 \] Let the first term be \( a \) and the common ratio be \( r \). The second term is \( ar \). So, \[ a - ar = 3 \] Using the sum formula: \[ \frac{a}{1 - r} = 27 \] From the equation \( a - ar = 3 \): \[ a(1 - r) = 3 \] Substituting \( a = \frac{27(1 - r)}{1 - r} = 27 \): \[ 27(1 - r) = 3 \] \[ 1 - r = \frac{3}{27} = \frac{1}{9} \] \[ r = 1 - \frac{1}{9} = \frac{8}{9} \] Thus, the correct answer is: \[ \frac{2}{3} \]