Need help with detail on proof regarding intermediate values of a derivative. This is Theorem 5.12 in Rudin's Principles of Mathematical Analysis.
Suppose $f$ is a real differentiable function on $[a,b]$ and suppose $f^{'}(a) < \lambda < f^{'}(b)$. Then there is a point $x \in (a,b)$ such that $f^{'}(x)=\lambda$.
Proof:
Put $g(t)=f(t)-\lambda \cdot t$. Then $g^{'}(a)<0$, so that $g(t_1) <g(a)$ for some $t_1 \in (a,b)$.
Can someone help me see why $t_1$ exists?
 A: $$g'(a)=\lim_{h\to0}\frac{g(a+h)-g(a)}{h}<0$$
implies that $g(a+h)-g(a)<0$ for sufficiently small $h>0$.
A: Suppose that $g(x)\geq g(a)$ for all $x\in[a,b]$. Then
$$
\frac{g(x)-g(a)}{x-a}\geq0\text{ for all }x\in(a,b],
$$
so that
$$
\liminf_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}\geq 0.
$$
Thus, it couldn't possibly be the case that $g'(a)<0$.
So, given that we know that $g'(a)<0$, it must be the case that $g(x)<g(a)$ for some $x\in(a,b]$.
A: Note that $$\lim_{x\to a} \frac{g(x)-g(a)}{x-a} = g'(a) < 0$$ This is basically right-handed derivative at $a$. Using the standard definition of limit, choose $\varepsilon = -\frac{g'(a)}{2}.$
Then $\exists\; \delta \gt 0$ such that $$\left|\frac{g(x)-g(a)}{x-a}-g'(a)\right| \lt -\frac{g'(a)}{2} \quad\forall\, x \in (a, a+ \delta)$$ $$\Rightarrow \frac{3}{2}g'(a) \lt g(x) - g(a) \lt \frac{g'(a)}{2}$$ Hence, $g(x)-g(a) \lt 0 \quad\forall\, x \in (a, a+ \delta)$. So there exists $t_1$ in $(a,b) \ni g(t_1) \lt g(a)$.
P.S: Similar procedure can be used for proving the existence of $t_2$ in the complete proof of the theorem.
