Applications of Derivative - Find no. of roots of

[AGNuke]

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 minimum number of zeroes of g(x) = (f'(x))²+f''(x)f(x) in the interval [a,e] is ...


All I can figure out is that at the least, it is a 4-degree polynomial with roots a, (b,c) (a root in between b and c), (c,d), e.

 

[Dick]

Here's a hint. What's the second derivative of f(x)^2?

 

[Simon Bridge]

f need not be a polynomial
f must have at least four roots. for eg. in the interval [b,c], f must cross the x-axis any odd number of times ;)

So sketch the least-zeros case - just guess.
For the same guess, sketch the first and second derivative. You'll see that f' is zero at turning points (and points of inflection - but you want least zeros, so your guess for f should probably avoid inflections) and f'' is zero when there are turning points in f'.

Notice that any number multiplied by zero is zero.
Compare your values with g(x).

 

[AGNuke]

This is my attempt. My mind was too fogged due to illness that I can't see something.

$$g(x)=\frac{\mathrm{d} }{\mathrm{d} x}(f(x).f'(x))$$

Integrating g(x) with respect to x.

$$G(x)=f(x).f'(x)$$

Now G(x) has minimum 7 zeroes (4 for f(x); 3 for f'(x)); therefore, g(x) must have minimum 6 zeroes.

 

[Simon Bridge]

That may not be the minimum for G(x) if f has a turning point at a zero.
This is part of why I suggested sketching f(x).

If you integrate an order 4 polynomial with 4 distinct zeros, do you usually get fewer zeros? Don't you normally reduce the order by differentiating?

In fact - looking at g(x) - it seems it may have trivial zeros where f'(x) and either f(x) or f''(x) are zero. ie either f(x) has a turning point where f(x)=0 or f'(x) has a turning point where f'(x)=0. Is this the case?

Non-trivially, (f'(x))^2 is always positive, while f''(x).f(x) may be positive or negative ... so there are likely to be some places where f''(x).f(x)=-(f'(x))^2

 

[AGNuke]

I sketched f(x). Roots of f(x) are a, (b,c), (c,d), e. I can think of a graph similar to a 4 degree polynomial.

I simply deduced G(x) which is ∫g(x).dx Looking upon the question, Since it is asking for minimum roots, f(x) is set to 4 degree polynomial, at least can be assumed for the given interval. We only have to find the roots of g(x) in [a,e], so 4 roots (minimum) of f(x) in the interval. So I think there should be no inflection points there.

As of f'(x), Since there are no inflection points, there must be (4-1)=3 zeroes, each between successive zeroes of f(x).

G(x) = f(x).f'(x) have 7 zeroes at minimum. Therefore, differential of G(x), which is g(x) have 6 zeroes.

I see nothing wrong in it...