Question

observe the given functions.

$f(x)=4x + 3$
$g(x)=(\frac{5}{3})^x$

complete the sentences to compare the two functions.

over the interval, the average rate of change of $g$ is greater than the average rate of change of $f$.
as the value of $x$ increases, the average rates of change of $f$ and $g$, respectively.
when the value of $x$ is equal to 7, the value of
it can be further generalized that a quantity increasing exponentially will exceed a quantity increasing linearly

Explanation:

Step1: Calculate average rate of change for \(f(x)\) and \(g(x)\)

The average rate of change of a function \(y = h(x)\) over the interval \([a,b]\) is \(\frac{h(b)-h(a)}{b - a}\).
For \(f(x)=4x + 3\), the average rate of change over any interval \([a,b]\) is \(\frac{(4b + 3)-(4a+3)}{b - a}=\frac{4(b - a)}{b - a}=4\) (constant).
For \(g(x)=(\frac{5}{3})^{x}\), over the interval \([4,6]\):
\(g(4)=(\frac{5}{3})^{4}=\frac{625}{81}\approx7.72\), \(g(6)=(\frac{5}{3})^{6}=\frac{15625}{729}\approx21.43\)
The average rate of change of \(g(x)\) over \([4,6]\) is \(\frac{g(6)-g(4)}{6 - 4}=\frac{\frac{15625}{729}-\frac{625}{81}}{2}=\frac{\frac{15625 - 5625}{729}}{2}=\frac{\frac{10000}{729}}{2}=\frac{5000}{729}\approx6.86>4\)

Step2: Analyze the trend of average - rate - of - change

For \(f(x)=4x + 3\), since it is a linear function \(y=mx + b\) with \(m = 4\), the average rate of change is constant (\(4\)).
For \(g(x)=(\frac{5}{3})^{x}\), as \(x\) increases, the average rate of change \(\frac{(\frac{5}{3})^{x_2}-(\frac{5}{3})^{x_1}}{x_2 - x_1}\) (where \(x_2>x_1\)) increases because the exponential function \(y = a^{x}\), \(a>1\) has an increasing slope.

Step3: Calculate function values at \(x = 7\)

\(f(7)=4\times7+3=28 + 3=31\)
\(g(7)=(\frac{5}{3})^{7}=\frac{78125}{2187}\approx35.73\)

Step4: Generalize the behavior of linear and exponential functions

For a linear function \(y=mx + b\) (\(m>0\)), it has a constant rate of change. For an exponential function \(y=a^{x}\), \(a > 1\), as \(x\) gets large enough, \(a^{x}\) will eventually exceed \(mx + b\) (by the property of exponential growth vs linear growth).

Answer:

Over the interval \([4,6]\), the average rate of change of \(g\) is greater than the average rate of change of \(f\). As the value of \(x\) increases, the average rates of change of \(f\) and \(g\) remain constant and increase, respectively. When the value of \(x\) is equal to \(7\), the value of \(g(x)\) is greater. It can be further generalized that a quantity increasing exponentially will eventually exceed a quantity increasing linearly.