To compare the growth rate of different functions, we can quantify the average rate of change (AROC) of each function

y = f (x)

over the interval

x_{2} - x_{1}

as shown below.

\underline{Average Rate of Change} \frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}

Since in general this rate changes along the function's graph, we should do this for several pairs of points to see how the rate is changing. Let's do this for each function so we can to compare them. We can start from the table of values of Exploration

2 ’ s

linear function.

$t$	$y = t$	$\frac{f ( x _{2} ) - f ( x _{1} )}{x _{2} - x _{1}}$	AROC
$1.0$	$1.0$	$\frac{2 . 0 - 1 . 0}{2 . 0 - 1 . 0}$	$1$
$2.0$	$2.0$	$\frac{2 . 0 - 1 . 0}{2 . 0 - 1 . 0}$	$1$
$3.0$	$3.0$	$\frac{4 . 0 - 3 . 0}{4 . 0 - 3 . 0}$	$1$
$4.0$	$4.0$	$\frac{4 . 0 - 3 . 0}{4 . 0 - 3 . 0}$	$1$

Since for a linear function the AROC is constant, we just calculated it for two pairs of points. Let's try now with the quadratic function.

$t$	$y = t^{2}$	$\frac{f ( t _{2} ) - f ( t _{1} )}{t _{2} - t _{1}}$	AROC
$1.0$	$1.0$	$\frac{4 . 0 - 1 . 0}{2 . 0 - 1 . 0}$	$3.0$
$2.0$	$4.0$	$\frac{4 . 0 - 1 . 0}{2 . 0 - 1 . 0}$	$3.0$
$3.0$	$9.0$	$\frac{1 6 . 0 - 9 . 0}{4 . 0 - 3 . 0}$	$7.0$
$4.0$	$16.0$	$\frac{1 6 . 0 - 9 . 0}{4 . 0 - 3 . 0}$	$7.0$
$5.0$	$25.0$	$\frac{3 6 . 0 - 2 5 . 0}{6 . 0 - 5 . 0}$	$11.0$
$6.0$	$36.0$	$\frac{3 6 . 0 - 2 5 . 0}{6 . 0 - 5 . 0}$	$11.0$
$7.0$	$49.0$	$\frac{6 4 . 0 - 4 9 . 0}{8 . 0 - 7 . 0}$	$15.0$
$8.0$	$64.0$	$\frac{6 4 . 0 - 4 9 . 0}{8 . 0 - 7 . 0}$	$15.0$

Finally, lets do the same for the exponential function.

$t$	$y = 2^{t} - 1$	$\frac{f ( t _{2} ) - f ( t _{1} )}{t _{2} - t _{1}}$	AROC
$1.0$	$1.0$	$\frac{3 . 0 - 1 . 0}{2 . 0 - 1 . 0}$	$2.0$
$2.0$	$3.0$	$\frac{3 . 0 - 1 . 0}{2 . 0 - 1 . 0}$	$2.0$
$3.0$	$7.0$	$\frac{1 7 . 0 - 7 . 0}{4 . 0 - 3 . 0}$	$8.0$
$4.0$	$15.0$	$\frac{1 7 . 0 - 7 . 0}{4 . 0 - 3 . 0}$	$8.0$
$5.0$	$31.0$	$\frac{6 3 . 0 - 3 1 . 0}{6 . 0 - 5 . 0}$	$32.0$
$6.0$	$63.0$	$\frac{6 3 . 0 - 3 1 . 0}{6 . 0 - 5 . 0}$	$32.0$
$7.0$	$127.0$	$\frac{2 5 5 . 0 - 1 2 7 . 0}{8 . 0 - 7 . 0}$	$128.0$
$8.0$	$255.0$	$\frac{2 5 5 . 0 - 1 2 7 . 0}{8 . 0 - 7 . 0}$	$128.0$

By comparing the last columns of the tables we get the following conclusions.

The AROC of the linear function is always the lowest.
Even though the exponential function's AROC is less than that of the quadratic function's one at first, the exponential function's AROC gets eventually greater, as it is increasing faster.
The quadratic function's AROC is increasing by a constant value, which is $4$ units each time. On the other hand, the exponential function's AROC is increasing by a constant factor, in this case it is being multiplied by $4$ each time.

Exercise