If f'(x)=(x-a)^2 n(x-b)^2 m+1 where m, n ∈ N, then

f'(x) = (x – a)²ⁿ (x – b)^{2m + 1}

∴  f'(x) = 0 ⇒ x = a, b

f'(x) does not change sign while passing through x = a. Hence 'a' is neither a point of maximum nor a point of minimum.