Roots Finding
f ( x ) = f ( 6 − x ) , f ′ ( 0 ) = f ′ ( 2 ) = f ′ ( 5 ) = 0 .
Consider all non-constant thrice differentiable functions f defined for all reals x that satisfies the above conditions. In the domain 0 ≤ x ≤ 6 , what is the minimum number of roots to
( f ′ ′ ( x ) ) 2 + f ′ ( x ) ⋅ f ′ ′ ′ ( x ) = 0 ?
Clarification : f ′ ( x ) , f ′ ′ ( x ) , f ′ ′ ′ ( x ) denote the 1 st , 2 nd , 3 rd derivatives of the function f ( x ) , respectively, with respect to x .
The answer is 12.
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1 solution
Moderator note:
You have only found a lower bound for the number of roots. You need to demonstrate that this is indeed the greatest lower bound, to conclude that it is the minimum.
E.g. At the start, we could say that f ′ ( x ) = 0 has 3 solutions, so f ′ ′ ( x ) = 0 has 2 solutions, so we arrive at the final answer of 3 + 2 − 1 = 4 . This demonsrates why the argument only provides a lower bound for the answer, as opposed to the greatest lower bound.
You have only found a lower bound for the number of roots. You need to demonstrate that this is indeed the greatest lower bound, to conclude that it is the minimum.
E.g. At the start, we could say that f ′ ( x ) = 0 has 3 solutions, so f ′ ′ ( x ) = 0 has 2 solutions, so we arrive at the final answer of 3 + 2 − 1 = 4 . This demonsrates why the argument only provides a lower bound for the answer, as opposed to the greatest lower bound.
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Yes, as indicated on my profile, I am part of the staff at Brilliant.
are you asking me or Calvin??
Reading absentmindedly, I have searched for the roots of f, and take f''(x)² + f'(x)f'''(x) = 0 as a constraint for all x. Maybe you should clarify that also
G i v e n f ( x ) = f ( 6 − x ) D . B . S . W . r . t x f ′ ( x ) = − f ′ ( 6 − x ) P u t x = 3 f ′ ( 3 ) = 0 P u t x = 0 f ′ ( 6 ) = 0 P u t x = 2 f ′ ( 4 ) = 0 P u t x = 5 f ′ ( 1 ) = 0 = > f ′ ( x ) = 0 h a s m i n i m u m 7 r o o t s i n 0 ≤ x ≤ 6 U s i n g r o l l e s t h e o r e m = > f ′ ′ ( x ) = 0 h a s m i n i m u m 6 r o o t s i n 0 ≤ x ≤ 6 c o n s i d e r a f u n c t i o n h ( x ) = f ′ ( x ) f ′ ′ ( x ) h ( x ) = 0 h a s m i n i m u m 1 3 r o o t s i n 0 ≤ x ≤ 6 h ′ ( x ) = ( f ′ ′ ( x ) ) 2 + f ′ ( x ) f ′ ′ ( x ) U s i n g r o l l e s t h e o r e m h ′ ( x ) = 0 h a s m i n i m u m 1 2 r o o t s i n 0 ≤ x ≤ 6