# Mean Value Theorem at infinity

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).

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 $$lambda < 0$$.

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.
• How do you know that $f(b - r) - f(a + r) \geq (b - a - 2r)\lambda$? Commented Nov 19, 2018 at 6:49
• @HD5450 I am using Mean Value Theorem. $f(b-r)-f(a-r)=f'(\xi) (b-a-2r)$ for some $\xi$ and my assumption is $f'(y) \geq \lambda$ for all $y$. Take $y=\xi$. Commented Nov 19, 2018 at 7:17
• How do we know that $f(b - r)$ and $f(a + r)$ are defined Commented Nov 19, 2018 at 7:21
• @HD5450 Your function id defined on $(a,b)$. $a+r$ and $b-r$ are points in this interval. Commented Nov 19, 2018 at 7:23