Given conditions on differentiability, what can we say about second derivative?

Question

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?

Answer 1

Hint

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

Answer 2

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 $x$.

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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.