# Minimum number of zeroes of $g(x)=(f'(x))^2+f''(x)f(x)$?

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 how many minimum number of zeroes does $g(x)=(f'(x))^2+f''(x)f(x)$ have in the interval $[a,e]?$

What way do I attempt a solution??

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Not sure whether this helps or not, but $g(x) = \frac{1}{2}(f(x)^2)''$. –  Michael Albanese Feb 17 '13 at 9:54
g(x)=1/2(f(x)^2)'', thanks for the information but how does it help? In the interval [a,e], the function has about 4 zeroes, hence if f is a polynomial, it must be, at least biquadratic. This means g is degree 6, but that does not help us to find number of zeroes in [a,e] –  Ashish Gaurav Feb 17 '13 at 10:02

Consider the function $h(x)=f(x)f'(x)$ and use Rolle's theorem.
@AshishGaurav: Note that $g(x)=h'(x)$. How many zeros does $h$ has? Hint:The function f has at least $4$ zeros and and $3$ relative extremes. Therefore ... –  P.. Feb 17 '13 at 9:59
$$g(x) = \frac{d}{dx} [f(x) f'(x)]$$
$f(x)$ has a minimum of $4$ zeroes in the given interval, while $f'(x)$ has $3$ in different places (between $a$ and $b$, $b$ and $c$, and $c$ and $d$). $f(x) f'(x)$ then has a minimum of $7$ zeroes in this interval, and consequently, its derivative $g(x)$ has $6$ zeroes at a minimum.