Krdytkyu

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 $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)>B$.

We say that $f(x)rightarrow -infty$ as $xrightarrow c$ if for every $Binmathbb{R}$ there is $delta>0$ so that $0<lvert x-crvert<deltaRightarrow f(x)<B$.

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 $lambda < 0$.

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 teh proof.