Say that we have function $f: \mathbb{R} \to \mathbb{R}$ and it is differentiable for all real values. Also, with $a,b \in \mathbb{R}$, the following is also true: $f'(x)=a$ for all $x$ and $f(0)=b$. Then, I am asked to find $f$ and prove that it is the only unique differentiable function with these properties.
I think it would be a good idea to use this proposition: If $f: [a,b] \to \mathbb{R}$ is differentiable, and for each $x \in [a,b], f'(x) = 0$, then $f$ is a constant function. So it seems that $f$ is $f(x) = ax + b$ where its derivative is constant for some $a$. But how can I utilize the $b$ to conclude that this is unique for $f$?
Best Answer
There is a "standard" trick to proving uniqueness in analysis. It consists in showing (or immediately using) that two functions $f$ and $g$ with certain properties have the same derivative, and thus differ only by an additive constant, and then showing that this constant is $0$.
In this specific example it goes like this: Let $g:\mathbb R\to\mathbb R$ be another differentiable function with $g'(x)=a$ for all $x\in\mathbb R$ and $g(0)=b$. Then $$(f-g)'(x)=f'(x)-g'(x)=a-a=0,$$ so $f-g$ is a constant function. But since $f(0)-g(0)=b-b=0$, it must be zero. So we actually have $f=g$.
This shows that any function satisfying your desired properties is equal to $f$.