Question

i need to find the rate of change for:
f(x)=0.01(2)^x
when x=2 and x=10
and for the table, the average rate of change of f(x) from x=1 to x=4

Answer 1

this is the table i referred to

Answer 2

rate of change from x = a to x = b\[\frac{ f(b)-f(a) }{ b-a }\]

applying this to your table:
rate of change from x = 1 to x = 4\[\frac{ f(4)-f(1) }{ 4-1}\]
you'll get your f(4) and f(1) values from the table

Answer 3

for the function - are you in calculus/learning about derivatives? I'm just a little confused as to whether it's asking for the instantaneous rate of change at x = 2 and x = 10, or the avg. rate of change between x = 2 and x = 10

Answer 4

so take the 16 and 2 to replace the top part of the equation and solve it?

Answer 5

^ yes, f(4) = 2 and f(1) = 16, and you'll plug these into the formula

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Mercury
for the function - are you in calculus/learning about derivatives? I'm just a little confused as to whether it's asking for the instantaneous rate of change at x = 2 and x = 10, or the avg. rate of change between x = 2 and x = 10
\(\color{#0cbb34}{\text{End of Quote}}\)
average rate of change, im in algebra 1

Answer 7

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Mercury
^ yes, f(4) = 2 and f(1) = 16, and you'll plug these into the formula
\(\color{#0cbb34}{\text{End of Quote}}\)
ok thank you

Answer 8

ok in that case, then for the function you'll plug in x = 2 and x = 10 then calculate
(f(10) - f(2) ) / (10-2)

(basically the same method as the formula)