How can this function possibly ever be greater than another? Say I have two functions, $f(x)$ and $g(x)$. I know that $f(x)' = 1$, and $g(x)' = (x + 1)^{-a}$, where a is some constant $0 < a < 1$. 
As you can see, $g'(x)$ is always less than or equal to one, and therefore always less than or equal to $f'(x)$ (to the right of the y-axis, of course). 
As far as I can tell, through integration, $f(x) = x$, and $g(x) = \frac{(x + 1)^{(1-a)}}{1-a}$. The problem here, is that for some values, $g(x) > f(x)$ in this case. How can this possibly be? 
If a function's derivative is less than some other function's at all points, how can that function ever be greater than it?
Example graph (I had to replace $f'(x)$ with $f(x)$, $g'(x)$ with $g(x)$, $f(x)$ with $h(x)$, and $g(x)$ with $j(x)$ due to desmos not supporting prime notation. 
 A: Note that if we have a derivable function $f$ then $f+c$ also has derivative $f'$.
A: You forgot about the integration constant.
$f(x) = x + C_1$ and $g(x) = \frac{(x+1)^{(1-a)}}{1-a} + C_2$

If a function's derivative is less than some other function's at all points, how can that function ever be greater than it?

There's nothing weird about that.
Also, $g(x)'$ is bigger than $1$ for some negative $x$
A: The value of the derivative of a function at any given point is not related in any way to the value of the function at that point. Antiderivatives are only defined up to a constant. What this means is that given any differentiable function $f(x)$ and any $c\in \mathbb{R}$, the function $f_c(x):=f(x)+c$ has derivative $f_c'(x)=f'(x)+0=f'(x)$. So "the" antiderivative of $f'(x)$ can take on any value you wish at any given point of $\mathbb{R}$. 
It sounds like maybe you're thinking about antidifferentiation as area accumulation, and reasoning that if we accumulate area under a smaller function then we get a smaller total area. This is true. If $g'(x)\leq f'(x)$, then $\int_0^x g'(t)\ dt \leq \int_0^x f'(t)\ dt$. The problem is that those integrals do not evaluate to $g(x)$ and $f(x)$, respectively, but rather $g(x)-g(0)$ and $f(x)-f(0)$. So you do obtain the inequality $g(x)-g(0)\leq f(x)-f(0)$. For your example functions, $f(0)=0<\frac{1}{1-a}=g(0)$, so you cannot deduce that $g(x)\leq f(x)$, but rather only $g(x)\leq f(x)+\frac{1}{1-a}$.
A: $g'(x)$ always less than $f'(x)$ implies (if the two are continuous) that $g(x)$ is always increasing at a lesser rate then $f(x)$ (or decreasing at a greater rate, or decreasing when the other is increasing or holding steady).
If, at some point, $f(x)$ is greater than $g(x)$ the difference is just going to get larger and larger as $x$ gets larger. (But also not that at earlier values of x the gap was less).
If, on the other hand, g(x) "starts" out larger at some point $x$, the gap is going to get smaller and smaller.  Likely $f$ and $g$ will intersect and afterwards $f$ will be larger the $g$.  If so, there is only one possible point of intersection and after $f$ will always be larger.
But also possible is that the gap gets smaller and smaller but never actually becomes zero.  A good example of the is $f(x) = 1 - 1/x$ on $(0, \infty)$ and $g(x) = 1$.  $f'(x) = 1/x^2 > 0 = g'(x)$.  $g(x)$ "begins" at 1 and stays at 1 forever.  $f(x)$ "starts" at "damned near -$\infty$" and spends "its entire life" trying to "catch up".
Your functions... well, you forgot the "plus a constant" refrain.  $f(x) = x + C_1$ and $g(x) = \frac{(x + 1)^{(1-a)}}{1-a} + C_2$.  The gap, $f(x) - g(x)$ between them is $(C_1 - C_2) + x -  \frac{(x + 1)^{(1-a)}}{1-a}$.  This gap is increasing.  If it "began" at a negative value when $g(x) > f(x)$ the gap will eventually increase to zero (and $f(x) = g(x)$ ["$f$ finally caught up"]) then increase to a positive value ($f(x) > g(x)$).
