Tangent lines of a differentiable function verifying $f(a) = f(b) = 0$ intersect the $x$ axis on all points outside of $[a, b]$.

Question

Let $f$ be a differentiable function on $[a, b]$, $(a<b)$ such as that $f(a) = f(b) = 0$.
Prove that $\forall d \in \mathbb{R} $\ $[a,b] $, $\exists c \in [a,b]$ such as the tangent line on $c$ intersects the $x$ axis on $(d,0)$.
My attempt: for arbitrary $d \in \mathbb{R} $\ $[a,b]$
Let $g_d$:x $\longmapsto f(x)(d-x)$, I thought this would work if I applied Rolle's Theorem to this function, until I derived it...
deriving it , we get $g_d'(x) = f'(x)(d-x) - f(x)$, I overlooked the minus sign...
So how should I proceed?

Answer 1

We want to prove that

$$\forall d\in \Bbb R-[a,b]\;\; \exists c\in[a,b]\; : $$ $$f'(c)(d-c)+f(c)=0$$

For this, define, for $ x\in[a,b]$

$$g(x)=\frac{f(x)}{d-x}$$

$g $ is continuous at $ [ a,b] $, differentiable at $ (a,b) $ and satisfies $$ g(a)=g(b)=0$$ then, by Rolle's Theorem

$$\exists c\in (a,b) \;\; : g'(c)=0$$ but for any $ x\in (a,b) $ $$g'(x)=\frac{f'(x)(d-x)-(-f(x))}{(d-x)^2}$$

Done.