Question

What is the average rate of change of the function f(x)=4(2)^x from x = 0 to x = 2?

Answer 1

late but there ya go also the other person was right :)

Explanation:

Answer 2

The average rate of change of the function
`f(x)` from
`x=0` to
`x=2` is equal to 6.

Explanation:

Let's define the average rate of change of the function
`f(x)` over the interval [a,b] :

`A(x)=` Δf / Δx =
`(f(b)-f(a))/(b-a)`

In the exercise,
`a=0` and
`b=2`

Now we calculate
`f(a)` and
`f(b)`

`f(a)=f(0)=4.(2^(0))=4.1=4`

`f(b)=f(2)=4.(2^(2))=(4).(4)=16`

⇒
`f(b)-f(a)=16-4=12` and
`b-a=2-0=2`

Finally, the average rate of change is

`(f(b)-f(a))/(b-a)=(12)/(2)=6`