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. 

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.
