I'm having trouble understanding the proof for the intermediate value theorem for derivatives. Here is the thereom from Rudin:
Suppose $f$ is a real differentiable function on $[a,b]$ and suppose $f'(x) < \lambda < f'(b)$. Then there is a point $x \in (a,b)$ such that $f'(x) = \lambda$.
Proof: Put $g(t) =f(t) - \lambda t$. Then $g'(a) < 0$, so that $g(t_1) < g(a)$ for some $t_1 \in (a, b)$, and $g'(b) > 0$, so that $g(t_2) < g(b)$ for some $t_2 \in (a, b)$. Hence $g$ attains its minimum on $[a, b]$ (Theorem 4.16) at some point $x$ such that $a < x < b$. By Theorem 5.8, $g'(x) = 0$. Hence $f '(x) = \lambda$.
I don't understand how $g'(a) < 0$ implies $g(t_1) < g(a)$ for some $t_1 \in (a, b)$. Same for $t_2$. Any help please?
Note that $$ \lim_{x\to a} \frac{g(x)-g(a)}{x-a} = g'(a) < 0 $$ implies that $$ \frac{g(x)-g(a)}{x-a} < 0 $$ for $x>a$ sufficiently close to $a$.