Proving the Relationship Between Roots of Two Equations

[tysonk]

If someone could guide me as to how I can approach this that would be appreciated. Suppose f(x) and g(x) have continuous first derivatives on R and that
f(x) g'(x) - g(x) f'(x) does not equal 0. Prove that between two consecutive roots of f(x) there is exactly one root of g(x).

 

[fzero]

Let

$$h(x) = f(x) g'(x) - g(x) f'(x) \neq 0$$

and let $$x_1,x_2$$ be consecutive roots of $$f(x)$$. Consider $$h(x_1),h(x_2)$$ and draw conclusions about the relative signs of $$g(x)$$ and $$f'(x)$$. You'll be able to determine something about the behavior of $$g(x)$$ on the interval $$(x_1,x_2)$$.

 

[tysonk]

Thanks for the reply.
So I concluded
h(x1) = -g(x1) f'(x1) =! 0
h(x2) = -g(x2) f'(x2) =! 0
This tells us that the derivative at the point of the root for f(x) must be either positive/negative. Also g(x) is not zero meaning there is no root for g(x) a those two points. However, I'm still having trouble understanding/concluding how there can be exactly one root for g(x) between the consecutive roots of f(x).

 

[fzero]

I would choose a definite sign for h(x), then you'll have to make similar choices for f' at the roots. This will lead conclusions for the sign of g at the roots. Having no root for f between $$x_1$$ and $$x_2$$ should put strong requirements on g and g'. You might also want to note that there will be a point between $$x_1$$ and $$x_2$$, where f' vanishes. That will lead to some more information on g.

 

[tysonk]

Thanks!