Problem with Rolle's theorem

Question

I really don't understand how to solve this. I have tried some solutions but they all assume that $f(a)=f(b)$. Can you give me more a more substantial hint?

Here is the problem

Answer 1

Well, the hint is not very subtle: you should look for a $\lambda$ such that $F(a)=F(b)$, id est...

Answer 2

Put

$$F(x)=f(x)-f(a)-\lambda (g(x)-g(a))$$

it is clear that $$F(a)=0$$

the constant $ \lambda $ will be chosen in a way to satisfy $$F(b)=0$$

$ F $ is continuous at $ [a,b ] $ , differentiable at $ (a,b) $ with $$ F(a)=F(b)$$

thus, by Rolle's Theorem,

$$(\exists c\in(a,b))\;\;:\;\; F'(c)=0$$

or $$f'(c)-\lambda g'(c)=0$$ with $$\lambda =\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$$

We are sure that $ g(b)-g(a)\ne 0$ because $ g'(x)\ne 0$ for any $ x\in(a,b)$.

Answer 3

The trick actually is hidden in how you construct $F(x)$.
Consider $ \dfrac{f(b)-f(a)}{g(b)-g(a)}= \text{constant}=\lambda $
Then it becomes $ \dfrac{f'(x)}{g'(x)}=\lambda $
Now integration yields $f(x)+\lambda g(x)=0=F(x)$