anonymous
  • anonymous
i need help plss. im not sure how to find the average rate. For f(x) = 0.01(2)x, find the average rate of change from x = 2 to x = 10. 1.275 8 10.2 10.24
Mathematics
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
the function is this? \[\LARGE f(x) = 0.01(2)^x\]
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
are you able to find the value of f(2) and f(10) ?

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jim_thompson5910
  • jim_thompson5910
let me know what you get for those values or let me know if you aren't sure how to find the function values
anonymous
  • anonymous
im not even sure how to find the function values.. i get all mixed up when it comes to trying to find the answers to these type of questions.
jim_thompson5910
  • jim_thompson5910
what you do is replace every x value with the input in question so say you wanted to find f(2) replace every x with 2 \[\LARGE f(\color{red}{x}) = 0.01(2)^\color{red}{x}\] \[\LARGE f(\color{red}{2}) = 0.01(2)^\color{red}{2}\] \[\LARGE f(2) = ???\]
jim_thompson5910
  • jim_thompson5910
at this point, you use a calculator to compute `0.01*(2)^2` you can use this calculator if you don't have one http://web2.0calc.com/
anonymous
  • anonymous
i got 0.04 @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
me too
jim_thompson5910
  • jim_thompson5910
so f(2) = 0.04
jim_thompson5910
  • jim_thompson5910
what is the value of f(10) ?
anonymous
  • anonymous
100,000,000 ? ._.
jim_thompson5910
  • jim_thompson5910
way too big
jim_thompson5910
  • jim_thompson5910
did you type in `0.01*(2)^10` ?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
which calculator are you using?
anonymous
  • anonymous
i got 10.24 this time.. i made a mistake
jim_thompson5910
  • jim_thompson5910
that's ok
jim_thompson5910
  • jim_thompson5910
now you compute \[\large \frac{f(b)-f(a)}{b-a} = \frac{f(10)-f(2)}{10-2} = \frac{10.24 - 0.04}{10-2} = ??\]
anonymous
  • anonymous
it equals 1.275
jim_thompson5910
  • jim_thompson5910
yep, that's your avg rate of change
anonymous
  • anonymous
thank you so much for your help! :)
jim_thompson5910
  • jim_thompson5910
you're welcome

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