Show that a function is surjective using Rolle's Theorem

Question

Let a,b ∈ $\mathbb{R}$, a < b and g: [a,b] → $\mathbb{R}$ is a continuous function, differentiable in ]a,b[, such that g(a) = g(b) = 0 and g(x) $\neq$ 0 $\forall$ x ∈ ]a,b[. Show that the function h:]a,b[ → $\mathbb{R}$,

h(x) = $\frac{g'(x)}{g(x)}$ is surjective.

$\forall$ k ∈ $\mathbb{R}$ apply, justifying, Rolle's Theorem to the function $g(x)e^{−kx}$ in [a,b]

I honestly have no idea where to start, or where the function $g(x)e^{−kx}$ comes from.

Answer 1

$h(x)=g(x)e^{-kx}$, $h(a)=h(b)=0$, there exists $c$ such that $h'(c)=(g'(c)-kg(c))^{-kc}=0$ (Rolle), we deduce that $g'(c)-kg(c)=0$ by dividing the previous equality by $g(c)$ (we are allowed to do that since $g(c)\neq 0$) we obtain ${{g'(c)}\over{g(c)}}=k$.

Answer 2

For any $k\in\Bbb R$, we want to find $x\in(a,b)$ such that $$h(x)=k$$$$\frac{g’(x)}{g(x)}=k$$$$g’(x)=kg(x)$$$$g’(x)e^{-kx}=kg(x)e^{-kx}$$$$(g’(x)-kg(x))e^{-kx}=0$$$$\frac{d}{dx}(g(x)e^{-kx})=0.$$ Since $g(a)e^{-ka}=g(b)e^{-kb}=0$, Rolle’s theorem guarantees a solution $x\in(a,b)$.