Prove there exists $m$ and $b$ such that $h(x) = mx + b$ Problem: 
Suppose ∃ function h: ℝ → ℝ such that h  as a second derivative h''(x) = 0 ∀ x ∈ ℝ. Prove ∃ numbers m, b:
h(x) = mx + b, ∀ x ∈ ℝ.

My attempt:
Consider h(x) = mx + b, with constants m and b. Note:
h(x) = mx + b
h'(x) = m + 0
h''(x) = 0. 
Hence the statement. 

This feels a little weak to me. I'd appreciate any suggestions on how better to prove (or if I'm totally off base, maybe set me on the right track). Thanks for any help in advance. 
 A: I am also not sure, but i would go the other way:
$h''(x) = 0 \Rightarrow \int h''(x) dx = h'(x)= m \Rightarrow \int h'(x) dx = h(x) =mx + b$
A: Because integration is the inverse of differentiation, we can find that if $h(x)=0,$
$$\int h ''(x) \,dx=\int 0\, dx\implies h'(x)=C_1\implies \int h'(x) \,dx=\int C_1 \,dx\implies h(x)=xC_1+C_2$$
Now we can choose $m=C_1$ and $b=C_2$, and we are done.
A: Solution 1 (with no integration). If $x\geq 0$, by the Mean Value Theorem (MVT) there exists $c_1>0$ such that
$$h(x)-h(0)=h'(c_1)x.\tag{1}$$
If $x<0$, again by the MVT, there exists $c_2>0$ such that
$$h(0)-h(x)=h'(c_2)(-x).\tag{2}$$
But, applying the MVT again, we conclude that
$$h'(c_1)-h'(c_2)=0(c_1-c_2)=0$$
and thus $h'(c_1)=h'(c_2)$. It follows from $(1)$ and $(2)$ that there exists $m=h'(c_1)$ and $b=h(0)$ such that, for any $x\in\mathbb{R}$,
$$h(x)=mx+b.$$
Solution 2 (with integration). By the Fundamental Theorem of Calculus (FTC),
$$h'(t)=h'(0)+\int_0^th''(x)\;dx=h'(0)+\int_0^t0\;dx=h'(0)+0=h'(0), \quad\forall\ t\in\mathbb{R}.$$
Take $x\in\mathbb{R}$. By the FTC again,
$$h(x)=h(0)+\int_0^x h'(t)\;dt=h(0)+\int_0^x h'(0)\;dt=h(0)+h'(0)\int_0^x 1\;dt=h(0)+h'(0)x$$
So, there exists $m=h'(0)$ and $b=h(0)$ such that $h(x)=mx+b$.
