Consider the function f(x) = ₃ ( k + 72 x ( 3 - x ) ( 3 + x ) 3x ) , where k is a real constant and x R . If the minimum value of f(x) is 2 , then the maximum value of f(x) is :

Step-by-step solution

Given function is: f(x) = ₃ ( k + 72 x ( 3 - x ) ( 3 + x ) 3x ) Using the identity x ( 3 - x ) ( 3 + x ) = 1 4 3x , the argument of the logarithm becomes: k + 72 ( 1 4 3x ) 3x = k + 18 3x 3x Using the double angle identity 2 = 2 : k + 18 3x 3x = k + 9 6x So, f(x) = ₃ (k + 9 6x) . Since the range of 6x is [-1, 1] , the minimum value of the argument is k - 9 and the maximum value is k + 9 . Given that the minimum value of f(x) is 2 : ₃ (k - 9) = 2 k - 9 = 3^2 = 9 k = 18 Now, we find the maximum value of f(x) : Maximu