If $f(x)$ is 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 find the minimum number of zeros of $g(x)=(f'(x))^2+f(x)f''(x)$in the interval $[a, e]$.
My attempt is as follows:-
By Intermediate value theorem, we can say following:
$1)$ There lies atleast one $c_1\in(b,c)$ such $f(c_1)=0$
$2)$ There lies atleast one $c_2\in(c,d)$ such that $f(c_2)=0$
$$g(x)=(f'(x))^2+f(x)f''(x)$$ $$g(x)=\dfrac{d(f(x)f'(x))}{dx}$$
Let, $h(x)=f(x)f'(x)$
$$h(a)=h(c_1)=h(c_2)=h(e)=0$$
There will be atleast three roots of $h'(x)$ by Rolle's theorem
By Rolle's theorem, we can also say following:-
$1)$ There lies atleast one $c_3\in(a,e)$ such that $f'(c_3)=0$
$2)$ There lies atleast one $c_4\in(b,d)$ such that $f'(c_4)=0$
As we have to find minimum no of roots, so it can be $c_1=c_3=c_4$
So $g(c_1)=f'(c_1)^2+f(c_1)f''(c_1)=0$, hence $c_1$ can be the root. We could have taken $c_2$ in place of $c_1$ but things would remain same.
So there should be minimum $4$ roots of $g(x)$. But actual answer is $6$. What am I missing here.
Rolle on $(a,c_1)$ gives $c_3$ such that $f'(c_3)=0$
Rolle on $(c_1,c_2)$ gives $c_4$ such that $f'(c_4)=0$
Rolle on $(c_2,e)$ gives $c_5$ such that $f'(c_5)=0$
We have $a<c_3<c_1<c_4<c_2<c_5<e$ and f or f' vanish at these points and also $h=ff'$
Rolle applied to $h$ on each interval gives 6 zéros of $g$.