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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

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  • $\begingroup$ Compute $g(7)-g(1)\over 7-1$. $\endgroup$ – David Mitra Sep 24 '12 at 21:19
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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 $g(7)$.

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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 $\dfrac{P(b)-P(a)}{b-a}$.

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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

$$ \begin{align} \frac{g(7) - g(3)}{7-3} &= \frac{3 + \frac{1}{2}\cdot 7 - (3 + \frac{1}{2}\cdot 3)}{7 - 3} \\ &= \dots \end{align} $$

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