Question

If f(x) is a twice differentiable function such that f(a) = 0, f(b) = 2, f(c) = −1, f(d) = 2, f(e) = 0, where a < b < c < d < e, then the minimum number of the zeroes of g(x) = (f′(x))² + f′′(x) f(x) in the interval [a, e] is _____. 

Answer 1

g(x) = d/dx(f(x)⋅f′(x)) 

To get the zero of g(x), we take function 

h(x) = f(x)⋅ f′(x)

Between any two roots of h(x), there lies at least one root of h′(x) = 0. That is,

g(x) = 0

Now,

h(x) = 0 and f(x) = 0

or f′(x) = 0 As f(x) = 0

has 4 minimum solutions and f′(x) = 0 has minimum 3 solutions, h(x) = 0 has minimum 7 solutions and h′(x) = g(x) = 0 has minimum 6 solutions.