Let f(x) = a x+1 + 4 4-x be a real valued function where a > 0 . If the maximum value of f(x) is 85 , then the minimum value of f(x) is

Step-by-step solution

The domain of the function is determined by x+1 0 and 4-x 0 , which gives x [-1, 4] . To find the maximum value of f(x) , we can use the Cauchy-Schwarz inequality: (a x+1 + 4 4-x )^2 (a^2 + 4^2)(( x+1 )^2 + ( 4-x )^2) (f(x))^2 (a^2 + 16)(x+1 + 4-x) (f(x))^2 5(a^2 + 16) Thus, the maximum value of f(x) is 5(a^2 + 16) . Given that the maximum value is 85 , we have: 5(a^2 + 16) = 85 5(a^2 + 16) = 85 a^2 + 16 = 17 a^2 = 1 Since a > 0 , we get a = 1 . Now, the function is f(x) = x+1 + 4 4-x . Since f(x) is the sum of two