Sorry for the broad title. The question is too long to fit in the title.
Here is the question:
Let $\space f,g: \space [a,b] \rightarrow \mathbb{R} \space$ be differentiable functions with $\space g'(x) \neq 0 \space$ for all $\space x\in (a,b).$
Suppose we define the following function:
$$h: \space [a,b] \rightarrow \mathbb{R}$$
With $\space h(x)=f(x)-kg(x)$.
Using Rolle's theorem for a suitable choice of $\space k \space$; prove that there exists a real number $\space c\in (a,b) \space$ such that:
$${f'(c)\over g'(c)} = {{f(b)-f(a)}\over {g(b)-g(a)}}$$
I gave it a try, but not sure if I'm right.
Here is what I did:
For Rolle's Theorem to be applied, we require $\space h(a)=h(b) \space $. In other words:
$$f(a)-kg(a) = f(b)-kg(b)$$
By rearranging, we get:
$$f(a)-f(b) = kg(a)-kg(b)$$
$$f(a)-f(b) = k \big(g(a)-g(b) \big)$$
$${{f(a)-f(b)}\over {g(a)-g(b)}} = {{f(b)-f(a)}\over {g(b)-g(a)}} = k$$
Which gives us the equality:
$${f'(c)\over g'(c)} = k$$
Here is where I get confused. Is this what it's asking me to do?
Or have I dived into the uncharted and gone in the completely wrong direction (won't be my first time haha).
Define $h(x)=f(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g(x)$. Then $h(b)=h(a)$ and so $h'(c)=0$ for some $c\in(a,b)$. But$$h'(c)=0\Longleftrightarrow\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$