Let f(x) = (x-2)^m (x-6)^n , where m, n 1, 2, , 10 . If f(x) attains a local minimum at an integer value (2, 6) , then the number of possible ordered pairs (m, n) is

Step-by-step solution

Given f(x) = (x-2)^m (x-6)^n . Differentiating with respect to x : f'(x) = m(x-2)^ m-1 (x-6)^n + n(x-2)^m(x-6)^ n-1 f'(x) = (x-2)^ m-1 (x-6)^ n-1 [m(x-6) + n(x-2)] f'(x) = (x-2)^ m-1 (x-6)^ n-1 [(m+n)x - (6m+2n)] The critical point in the interval (2, 6) is: = 6m+2n m+n = 2 + 4m m+n Since is an integer and (2, 6) , the possible values for are 3, 4, and 5 . Case 1: = 3 2 + 4m m+n = 3 m+n = 4m n = 3m Case 2: = 4 2 + 4m m+n = 4 2m+2n = 4m m = n Case 3: = 5 2 + 4m m+n = 5 3m+3n = 4m m = 3n For f(x) to have a local mini