Let f : R R be a thrice differentiable function such that f(0) = 0, f(1) = 1, f(2) = -1, f(3) = 2 and f(4) = -2. Then, the minimum number of zeros of (3f'f'' + ff''')(x) is .
Answer
5
Solution
(f2)' = 2ff', (f2)'' = 2(f')2 + 2ff''
(f2)''' = 6f'f'' + 2ff''' = 2(3f'f'' + ff''')
⇒ zeros of 3f'f'' + ff''' = zeros of (f2)'''
f: zero at x = 0 and sign changes in (1, 2), (2, 3), (3, 4) ⇒ at least 4 zeros
(f2)' = 2ff': these 4 zeros of f + 3 zeros of f' between them (Rolle) ⇒ at least 7 zeros
Rolle's theorem each time: (f2)'' has at least 6, (f2)''' at least 5 zeros
Answer: 5