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Mathematics 20 Online
OpenStudy (anonymous):

Which of the following options results in a graph that shows exponential decay? f(x) = 0.6(2)x f(x) = 3(0.7)x f(x) = 0.4(1.6)x f(x) = 20(3)x

jimthompson5910 (jim_thompson5910):

In general, the exponential function is \[\Large f(x) = a*b^x\] The base b determines if you have exponential growth or decay. If `0 < b < 1`, then you have exponential decay. If `b > 1`, then you have exponential growth.

jimthompson5910 (jim_thompson5910):

For example, the function \[\Large f(x) = 33(2.7)^x\] has exponential growth because b = 2.7 is larger than 1.

OpenStudy (anonymous):

Ok so for example f(x) = 3(0.7)to the x power is exponential decay @jim_thompson5910

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

since b = 0.7 is between 0 and 1 ie it makes `0 < b < 1` true

OpenStudy (anonymous):

Ok thank you

jimthompson5910 (jim_thompson5910):

np

OpenStudy (anonymous):

Hey wait can you help me solve another problem? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

sure

jimthompson5910 (jim_thompson5910):

go ahead

OpenStudy (anonymous):

For f(x) = 0.01(2)to the x power, find the average rate of change from x=2 to x=10

jimthompson5910 (jim_thompson5910):

are you able to determine f(2) ?

OpenStudy (anonymous):

Nope that's all the question says

jimthompson5910 (jim_thompson5910):

replace every x with 2 and evaluate the function

OpenStudy (anonymous):

These are the choices: 1.275 8 10.2 10.24

jimthompson5910 (jim_thompson5910):

ignore the choices for now

jimthompson5910 (jim_thompson5910):

\[\Large f(x) = 0.01(2)^x\] \[\Large f(2) = 0.01(2)^2\] \[\Large f(2) = ??\]

OpenStudy (anonymous):

0.04

jimthompson5910 (jim_thompson5910):

yes

jimthompson5910 (jim_thompson5910):

now evaluate f(10)

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

I got 1,024

jimthompson5910 (jim_thompson5910):

you should get something smaller than that, try again

jimthompson5910 (jim_thompson5910):

\[\Large f(x) = 0.01(2)^x\] \[\Large f(10) = 0.01(2)^{10}\] \[\Large f(10) = ??\]

OpenStudy (anonymous):

10.24

jimthompson5910 (jim_thompson5910):

better

OpenStudy (anonymous):

Is that right

jimthompson5910 (jim_thompson5910):

Last step: compute \[\Large \frac{f(b)-f(a)}{b-a}=\frac{f(10)-f(2)}{10-2} = ??\]

jimthompson5910 (jim_thompson5910):

10.24 is not the answer, but it helps get you there

OpenStudy (anonymous):

Wait is it 8over 8

jimthompson5910 (jim_thompson5910):

\[\Large \frac{f(10)-f(2)}{10-2}=\frac{10.24-0.04}{10-2} = ??\]

OpenStudy (anonymous):

Oh ok

OpenStudy (anonymous):

So it 1.275

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (anonymous):

And that's the right answer

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (anonymous):

Ok thank you

jimthompson5910 (jim_thompson5910):

np

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