Determining the average rate of change? Yeah, need help solving this one...
A function is given below. Determine the average rate of change of the function between x = 1 and x = 7. 
g(x)= 3 + 1/2x
 A: Suppose that $g(1)=4$ and $g(7)=2$. (Neither of these is true, but I’ll use them to illustrate what’s going on.) Then the function has changed by $2-4=-2$ units between $x=1$ and $x=7$. That’s $-2$ units of change in $g$ while $x$ changed by $7-1=6$ units, so on average $g$ changed by $\frac{-2}6=-\frac13$ of a unit every time $x$ increased by $1$ unit. That is, the average rate of change was $-\frac13$ (units change in $g$ per $1$-unit change in $x$).
Now you try it with the actual values of $g(1)$ and $g(7)$.
A: A particle is travelling in a straight line. Suppose that $f(t)$ is the displacement (position) of the particle at time $x$. Then the average velocity from time $t=a$ to time $t=b$ is $\dfrac{f(b)-f(a)}{b-a}$. It is the change in displacement, divided by the time it took.  In other words, it is the average rate of change of displacement.
If the population of a country at time $t$ is $P(t)$, then the average rate of change in population from time $t=a$ to time $t=b$ is the change in population, divided by the time it took. This is $\dfrac{P(b)-P(a)}{b-a}$.
A: Saying what others have said a tiny bit differently: By definition the average rate of change of a function $g$ over the interval $[a,b]$ is
$$
\frac{g(b) - g(a)}{b-a}.
$$
In you case $g(x) = 3 + \frac{1}{2}x$ and $a = 1$ and $b = 7$. So you have
$$
\begin{align}
\frac{g(7) - g(3)}{7-3} &= \frac{3 + \frac{1}{2}\cdot 7 - (3 + \frac{1}{2}\cdot 3)}{7 - 3} \\
&= \dots
\end{align}
$$
