Question

For a twice differentiable function f(x), g(x) is defined as g(x) = (f'(x))² + f''(x) f(x) on [a, e]. If a < b < c < d < e, f(a) = 0, f(b) = 2, f(c) = – 1, f(d) = 2, f(e) = 0, then find the maximum number of zeroes of g(x).

Answer 1

We have g(x) = (f'(x))² + g'(x)f''(x)

g(x) = d/dx (f(x).f'(x))

By the Rolle’s theorem, between any two roots of a polynomial, there is a root of its derivative.

Now, f(x).f'(x) = 0

Either f(x) = 0 or f'(x) = 0

Thus, f(x) has four zeroes at x = a, between b and c, between c and d and at e

So, f'(x) has atleast 3 zeroes.

Thus, f(x)f'(x) has atleast 7 zeroes.

Therefore, g(x) has atleast 6 zeroes.