Question

If f(x) is a twice differentiable function for all x belonging to R such that f(0) = 0, f(2-x) = f(2+x), f(4-x) = f(4+x), f(6-x) = f(6+x), .... then what is the minimum number of zeros of g(x) = f(x)f''(x) + (f'(x))^2 in the interval [0, 13] ?

Answer 1

g(x) = d/dx( f(x) * f'(x) )
Hint: Rolle's Theorem.

Answer 2

In the interval [0,13], we know for sure f(x) attains zero at least at one point, at x = 0.

f(2-x) = f(2+x) implies, by Rolle's Theorem, that there is a 'c' in the interval (2-x, 2+x) where f'(c) = 0. If we make x smaller and smaller we can conclude f'(2) = 0.

So in the interval [0,13] we know for sure f'(2) = f'(4) = f'(6) = f'(8) = f'(10) = f'(12) = 0.

Let h(x) = f(x) * f'(x)
Then h'(x) = f(x) * f''(x) + { f'(x) }^2 = g(x)

h(x) = f(x) * f'(x)
h(x) is zero when either f(x) = 0 or f'(x) = 0 (or both)
In the interval [0,13], we know for sure, f(x) is zero at least at one point (at x = 0) and f'(x) is zero, at least at 6 points (at x = 2, 4, 6, 8, 10, 12).
Therefore, in the interval [0,13], h(x) becomes zero at least at 7 points (at x = 0, 2, 4, 6, 8, 10, 12).
h(0) = h(2) = h(4) = h(6)= h(8) = h(10) = h(12) = 0 (7 points)
Using Rolle's theorem again, there must be at least 6 points in the interval [0,13] where h'(x) = 0 { in the intervals: (0,2), (2,4), (4,6), (6,8), (8,10), (10,12) }

Therefore, g(x) (= h'(x)) must have at least 6 roots in the interval [0,13].

Answer 3

Were you trying to challenge us?

Answer 4

Came across this interesting question and I thought I should let others know.

Answer 5

Cool, thanks for sharing :). Brings back old memories of Rolle's Theorem.