The polynomial at) 2 23:5 935 + 153:4 10e3 + 33: 1 has a stationary point at a: = 1. This is because f(1)(1) = 0 0 Calculate the higher derivatives: . f(2)(1) = 0 o . f(3)(1) = 0 o . f(4)(1) = 0 o .f(5)(1) = -. So the smallest positive integer 'n, > 1 for which f(n) (1) 75 0 is Hence the function has a horizontal point ofinfl... 0 at :1: = 1