Question

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

Enter your answer in the box.

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Answer 1

@Vocaloid

Answer 2

if you didn't understand something about moreso's work, you are welcome to ask questions

Answer 3

i didn't know how to do it and it's better if you teach me

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @moreso
\(\color{#0cbb34}{\text{Originally Posted by}}\) @moreso
\(\color{#0cbb34}{\text{Originally Posted by}}\) @moreso
use the formula \[ \Large f_{avg}= \frac{f(b)-f(a)}{b-a}\]
\(\color{#0cbb34}{\text{End of Quote}}\)
\(\color{#0cbb34}{\text{End of Quote}}\)
\(\color{#0cbb34}{\text{End of Quote}}\)


in this case, b is your "upper" x value (2) and a is your lower x value (0)

so let b = 2 and a = 0 and calculate the rate using the above formula

Answer 5

i just calculated it and its 40 ?

Answer 6

thats incorrect

Answer 7

\[ \large f_{avg}= \frac{ 4(2)^2-4(2)^0}{2-0}\]

Answer 8

i kind of need to get work done tonight, I'll let moreso take over ^_^ thanks

Answer 9

\[ \large f_{avg}= \frac{ 4(2)^2-4(2)^0}{2-0} = \frac{4\times 4 - 4 \times 1 }{2} \]

Answer 10

14

Answer 11

(16-4)/2

Answer 12

6

Answer 13

it's 6

Answer 14

correct me if i'm wrong

Answer 15

yes 6

Answer 16

ok thanks moreso and vocaloid