Mean Value Theorem at infinity

Question

Suppose $f: (a,b) \rightarrow \mathbb{R}$ is differentiable with $\lim_{x \rightarrow a} f(x) = \infty$ and $\lim_{x \rightarrow b} f(x) = -\infty$. Suppose also that there are $c_1 < c_2 \in (a, b)$ with $f(c_1) \leq f(c_2)$. For any $\lambda < 0$, there is $x_0 \in (a, b)$ with $f'(x_0) = \lambda$.

We say that $f(x)\rightarrow \infty$ as $x\rightarrow c$ if for every $B\in\mathbb{R}$ there is $\delta>0$ so that $0<\lvert x-c\rvert<\delta\Rightarrow f(x)>B$.

We say that $f(x)\rightarrow -\infty$ as $x\rightarrow c$ if for every $B\in\mathbb{R}$ there is $\delta>0$ so that $0<\lvert x-c\rvert<\delta\Rightarrow f(x)<B$.

My attempt: so my idea for the problem is that I can just consider halfof the graph and pick a point $\alpha$ that is nearby to $a$, then consider the interval $[\alpha, c_2]$, we can get the slope with the slope formula and we get $\frac{f(\alpha) - f(c_2)}{\alpha - c_2} > \frac{B - f(c_2)}{\alpha - c_2} \geq \lambda + 1 > \lambda$.

My question: I am not sure how to pick a point that is nearby to $a$, will I be picking a point that is within the delta neighbourhood? Also, I don't know how to prove $lambda < 0$.

Answer 1

All derivatives have IVP. So it is enough to show that there exists points $x,y$ with $f'(x) >\lambda$ and $f'(y) <\lambda$. If $f'<0$ at every point we have a contradiction to the hypothesis that $f(c_1) \leq f(c_2)$. So $f' \geq 0 >\lambda$ at some point. Now suppose $f'(y) \geq \lambda$ for all $y$. Then $f(b-r)-f(a+r) \geq (b-a-2r)\lambda$ for all $r>0$ sufficiently small (by MVT). Letting $r \to 0$ we a get contradiction to the fact that $f(b-)=-\infty$ and $f(a+)=\infty$. This completes the proof.