1. ## functions

let me ask you this one..

we got f(x), g(x), in both there is differentiation.

for all x, we know: $f'(x)*g(x)$ $g'(x)*f(x)$

we also know there are: $f(a)=f(b)=0$

we need to prove that there is a point 'c', in zone [a,b], which gives us g(c)=0.

2. Hi

Suppose there exists no c in zone [a,b], which gives us g(c)=0

Consider $h(x) = \frac{f(x)}{g(x)}$

h is therefore defined and differentiable for all x in [a,b] ...

3. firstly thanks for your response.

let's see if i got it right.

based on Rolle, we say there is $f'(d)=0$ on [a,b]

then we say, $h(x)=f(x)/g(x) , h(x)$ $f'(x)/g'(x)$

but this could not be, because we have $f(a)=0 , f'(d)=0$ both in zone [a,b]

am i right ??

4. Originally Posted by dannee

let's see if i got it right.

based on Rolle, we say there is $f'(d)=0$ on [a,b]

then we say, $h(x)=f(x)/g(x) , h(x)$ $f'(x)/g'(x)$

but this could not be, because we have $f(a)=0 , f'(d)=0$ both in zone [a,b]

am i right ??
sorry i don't follow ur argument. But here is another hint adding to what running-gag said, if $h(x) = f(x)/g(x)$ then $h'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g(x)^2}$ what does the numerator remind you of?

5. it means h'(x) is a monotonic function, but i don't know how to go from there

6. Originally Posted by dannee
it means h'(x) is a monotonic function, but i don't know how to go from there
h(a)=0=h(b) use a special theorem to derive a contradiction.

7. got it, thank you very much