Assume that $f : A \rightarrow \mathbb{R}$ is two times differentiable with $f''(x)=0$ for all $x \in A$, with $A$ an interval.
Show, (not by integration), that $f$ is of the form $f(x) = ax + b$ for some constants $a, b$.
My answer: I will use the Mean Value Theorem:
Take $x,y \in A, x
. Applying the Mean Value Theorem on $[x,y]$ gives: $f''(c)=\frac{f'(y)-f'(x)}{y-x}$ for some $c \in A$. If $f''(x)=0 $ for all $x \in A$, then $f''(c)=0$, wich means that $f''(c)=\frac{f'(y)-f'(x)}{y-x}= 0$, which implies that $f'(y)-f'(x)=0$. So $f'(y)=f'(x)$. Set $k$ equal to this common value. Because x and y are arbitrary, it follows that $f'(x)=a$ for all $x \in A$ We are now given that $f'(x)=a$ for all $x \in A$. This means that, if we take $x,y \in A, x
, by applying the Mean Value Theorem on [x,y], $f'(c)=\frac{f(y)-f(x)}{y-x}$ for some $c \in A$. We know that $f'(x)=a$ for all $x \in A$, which means that $f'(c)=\frac{f(y)-f(x)}{y-x}=a \implies f(y)-f(x)=a(y-x)$
What's next?
I could also use the definition of derivative, considering:
$f'(x)=a = \lim_{x \to c} \frac{f(x)-f(c)}{x-c}$
How can I show that $f(x) = ax + b$ ?