Given conditions on differentiability, what can we say about second derivative? Let $f:[0,10] \rightarrow [10,30]$ be a continuous and twice differentiable function such that $f(0)=10$ and $f(10) =20$. Suppose $\mid f'(x) \mid \leqslant 1 \; \forall x \in [0,10]$. Then the value of $f"(5)$ is
$(a) 0$
$(b) 1/2$
$(c) 1$
$(d) Can't be determined$
My trial:
We need the function to be twice differentiable. So let $f"(x)=c$ where $c$ is a constant.
Then $f(x) =\frac{cx^{2}}{2} + dx + e$ where $c,d,e \in \mathbb{R}$
Also, $f(0) =10$ implies $e=10$
and $f(10) = 20 \;$ implies $ \; 50c + 10d =10$ or $5c + d=1$
Now, $f'(5) = 5c +d = 1$ (suprise, shawty!)
therefore $c = \frac{1-d}{5} , d \in \mathbb{R}$
How to proceed? what values of $d$ to chooses such that $\mid f'(x) \mid \leqslant 1$
Any other easier approach is also welcome. Thanks. Can we somehow use MVT or Brower's Fixed Point here?
 A: You cannot assume that $f''$ is constant (though it will turn out that it is).
Assume $f'(a)<1$ for some $a\in[0,1]$. Then by continuity(!) of $f'$, there exists an interval $[a,b]$ and a number $q<1$ with $f'(x)<q$ for all $x\in[a,b]$.
Then
$$ \begin{align}10&=f(10)-f(0)\\
&=\int_0^{10}f'(x)\,\mathrm dx \\
&=\int_0^{a}f'(x)\,\mathrm dx+\int_a^bf'(x)\,\mathrm dx+\int_b^{10}f'(x)\,\mathrm dx\\
&\le \int_0^{a}1\,\mathrm dx+\int_a^bq\,\mathrm dx+\int_b^{10}1\,\mathrm dx\\
&=a+(b-a)q+10-b\\&=10-(b-a)(1-q)\\
&<10,\end{align}$$
contradiction.
Therefore $f'(x)=1$ for all $x$.

In fact, let's forget about the second derivative, i.e., start with once differentiable $f\colon [0,10]\to [20,30]$ with $f(0)=20$, $f(10)=30$, and $f'(x)\le 1$ for all $x$. Then this is enough to show $f(x)=x+20$ (and consequently, $f''$ exists and is identically $0$):
Consider $x$ with $0<x<10$

*

*Assume  $f(x)>x+20$. Then by the Mean Value Theorem, there exists $\xi$ between $0$ and $x$ with
$$f'(\xi)=\frac{f(x)-f(0)}{x-0}>\frac{(x+20)-(0+20)}{x} =1,$$
contradiction.

*Assume $f(x)<x+20$. Again by the mean value theorem, there eixts $\xi$ betwewn $x$ and $10$ with
$$f'(\xi)=\frac{f(10)-f(x)}{10-x}>\frac{30-(x+20)}{10-x} =1,$$
contradiction.

*Therefore, the only remaining possibility $f(x)=x+20$ remains.

A: Hint
One can prove that $f'(x)=1$ for all $x\in [0,10]$.
