Let f(x) = (x-1)^a (x-4)^b , where a and b are positive integers. It is given that f(x) attains a local maximum at x=2 , and local minima at both x=1 and x=4 . The minimum possible value of a+b is

Step-by-step solution

Given f(x) = (x-1)^a (x-4)^b . Differentiating with respect to x : f'(x) = a(x-1)^ a-1 (x-4)^b + b(x-1)^a(x-4)^ b-1 f'(x) = (x-1)^ a-1 (x-4)^ b-1 [a(x-4) + b(x-1)] f'(x) = (x-1)^ a-1 (x-4)^ b-1 [(a+b)x - (4a+b)] The critical point in the interval (1, 4) is obtained by setting the linear factor to zero: x = 4a+b a+b Since f(x) attains a local maximum at x=2 : 4a+b a+b = 2 4a+b = 2a+2b b = 2a For f(x) to attain local minima at x=1 and x=4 , the function must not change sign at these roots. Since f(1) = 0 and f(4) = 0