Question

the function ( f(x) ), shown in the graph, represents an exponential growth function. compare the average rate of change of ( f(x) ) to the average rate of change of the exponential growth function ( g(x)=29(1.4)^{x} ). for both functions, use the interval ( 0,4 ).
on the interval ( 0,4 ), the average rate of change for ( f(x) ) is ( square ) and the average rate of change for ( g(x) ) is ( square ). so, ( square ) has the greater average rate of change on the given interval. (round to the nearest tenth as needed.)

Explanation:

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

The formula for the average rate of change of a function \( y = f(x) \) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\).
For \( f(x) \), \( a = 0\), \( f(0)=20\), \( b = 4\), \( f(4)=45.2\).

$$ \frac{f(4)-f(0)}{4 - 0}=\frac{45.2-20}{4}=\frac{25.2}{4}=6.3 $$

Step2: Calculate the average rate of change for \( g(x) \)

For \( g(x)=29(1.4)^{x}\), \( a = 0\), \( g(0)=29(1.4)^{0}=29\), \( b = 4\), \( g(4)=29(1.4)^{4}\).
First, calculate \( (1.4)^{4}=1.4\times1.4\times1.4\times1.4 = 3.8416\).
Then \( g(4)=29\times3.8416 = 111.4064\).

$$ \frac{g(4)-g(0)}{4 - 0}=\frac{111.4064 - 29}{4}=\frac{82.4064}{4}=20.6016\approx20.6 $$

Answer:

On the interval \([0,4]\), the average rate of change for \( f(x) \) is \(6.3\) and the average rate of change for \( g(x) \) is \(20.6\). So, \(g(x)\) has the greater average rate of change on the given interval.