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let $f:[0,10]\to[10,20]$ be a continuous and twice differentiable function such that $f(0)=10$ and $f(10)=20$. Suppose $|f'(x)| \leq 1$ for all $x \in[0,10]$. Then, the value of $f"(5)$ is?

This is my crude way of doing it

I thought of it like this, the maximum function that can increase is with slope 1. So if it increases anywhere at less than slope 1 then at some other point it has to increase with a slope greater than 1 for it to reach the point (10,20) so it has to be linear fun. with slope 1 everywhere and therefore f"(x)=0 but can someone give me some formal proof of it?

insipidintegrator
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Siddharth Tyagi
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1 Answers1

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$20=f(10)=f(0)+\int_0^{10}f'(t)dt\leq 10+10=20$ since $f'(t) \leq 1$ for all $t$. This inequality forces $f'(t)$ to be $1$ for all $t$. [If $f'(t)<1$ for some $t$ then $f'(t)<1$ in some open interval contained in $[0,10]$ and this forces $\int_0^{1} f'(t)dt$ to be strictly smaller than $10$]. Hence, $f'(5)=1$. Also, $f(5)=f(0)+\int_0^{5} 1dt=15$.

geetha290krm
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