If $f' (x) = c$, a constant, for all $x$, show that $f(x) = cx + d$ for some constant $d$. If $f' (x) = c$, a constant, for all $x$, show that $f(x) = cx +
d$ for some constant $d$. 
By the limit definition $\lim_{x_0 \to 0} \frac{f(x+x_0)-f(x_0)}{x_0} = c$.
How can I get that the function is linear? I don't know about integrals yet and can't use them. 
 A: Consider $g(x)=f(x)-cx$. Then $g’(x)=f’(x)-c=c-c=0$.
Therefore $g$ is constant, say $g(x)=d$.
Then $f(x)=cx+d$.
A: Suppose that $F(x)$ is an antiderivative of $f'(x)$. We know that $F(x)$ and $f(x)$ differ by only a constant. Therefore:
$$f(x) = F(x) + d$$
for some constant $d$.
Take the derivative of $cx$. If you get $c$, then you know that $F(x) = cx$ is an antiderivative of $f'(x)$. Thus, you have $f(x) = F(x)+d = cx+d$.
This uses the definition of antiderivative, not but does not actually calculate an integral. If you have not yet learned that any two antiderivatives of a function may differ by at most a constant, then this answer is not sufficient, of course. The proof of this involves the Mean Value Theorem. So, any answer will depend greatly on what you have already learned.
If you must use the Mean Value Theorem, then suppose you have two functions $F(x), G(x)$ such that $F'(x) = G'(x) = f'(x)$. In other words, $F(x)$ and $G(x)$ are both antiderivatives of $f'(x)$.
Define $H(x) = F(x) - G(x)$.
Then $H'(x) = F'(x) - G'(x) = f'(x)-f'(x) = 0$ is identically zero for all $x$.
Let $a<b$ be any arbitrary real numbers. Then by the Mean Value Theorem, there exists $c \in (a,b)$ such that $H(b)-H(a) = H'(c)(b-a) = 0(b-a) = 0$ shows that $H(b)-H(a)=0$ for all $a,b$, so $H$ is a constant function.
Thus, any two antiderivatives of a function differ by at most a constant.
