Find the differential function $f$ satisfies the following conditions

Question

Find the differential function $f:\mathbb{R}\to\mathbb{R}$ satisfies the following conditions:

(1) $1\le f'(x)\le 2$ for all $x\in\mathbb{R}$,

(2) $f(1)=2$ and $f(2)=3$.

I guess the answer is $f(x)=x+1$ by M.V.T. or Rolle's theorem.

So, I tried with M.V.T and Rolle's theorem, but can't prove it rigorously.

Can anyone help me? Thank you!

Answer 1

Define $ g(x)=x+1.$

Note that $f'(x)\ge g'(x)=1$ and $f(1)=g(1)=2.$

Therefore $f(x)\ge g(x)$ for $ x\ge 1$

Since $g(2)=f(2)$, we realize that $f'(x)$ does not exceed $g'(x)$ at any point between $1$ and $2$.

Therefore $ f'(x)=g'(x)=1$

Thus $f(x) =x+C$.

From $f(1)=2$ we find $C=1$, that is $f(x)=x+1.$

Answer 2

$1=3-2=f(2)-f(1)=\int _1 ^{2} f'(x)dx \geq \int _1 ^{2} 1dx =1$. This is possible only when $f'(x)=1$ for all x (because if $f'(a)>1$ for some a then the inequality holds in some interval around a which makes the integral of $f'$ greater than 1). Now $f'(x)=1$ for all x implies $f(x)-f(1)=(x-1)$ by MVT. Hence $f(x)=f(1)+x-1=x+1$. Note that the condition $f'(x) \leq 2$ is not needed.