Question

Find the relative rate of change fro each of the following functions.
f(x)=x^3 -3x^2

Answer 1

I think you have to derive here, or am I wrong? Since it's about rate of change, right?

Answer 2

no this is the problem

Answer 3

the derivative would be 3x^2 -6x

Answer 4

exactly

Answer 5

that's how you compute the relative rate of change, by deriving ^_^

Answer 6

im not understanding?

Answer 7

so do i take f'(x) and divide it by f(x)

Answer 8

lol, to find the relative rate of change, you have to find the derivative of the following function which is in this case :\[f(x) = x^3 - 3x^2\]


so when you derive the function you'll get like what you've said :
\[f'(x) = 3x^2 - 6x\]

since there is no given for x, then that's the final equation I guess :)

Answer 9

if they told you, for example, that x = 2 then substitute it in the derived function and you'll find the relative rate of change

Answer 10

^_^ clearer?

Answer 11

or am I mixing it up with something?

Answer 12

i got all that, but my teacher still counted 2 points off?

Answer 13

hmm, I'm missing a point aren't I?

Answer 14

sstarica, when you're done here, can you help dinaortega? I'm leaving soon.

Answer 15

I'll try my best loki :)

Answer 16

could you help me lokisan?

Answer 17

Thanks.

Answer 18

np

Answer 19

relative rate of change is f'(x)/f(x)

Answer 20

>_< thank you blexting

Answer 21

blexting so how do i get that from f(x)= x^3 -3x^2 being the original problem and f'(x)=3x^2 -6x

Answer 22

Sorry.. I don't know, but I knew the formula.. sstarica can you help?

Answer 23

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Answer 24

divide f'(x) with f(x)

Answer 25

and simplify

Answer 26

x^3-6x?

Answer 27

6x/x^3

Answer 28

\[\frac{f'(x)}{f(x)}=\frac{3x^2-6x}{x^3-3x^2}=\frac{3x-6}{x^2-3x}\]

Answer 29

so you'll get :\[f'(x)/f(x) = (3x^2-6x)/(x^3-2x^2)\]\[= 3x(x-2)/x^2(x-2)\]

\[= 3/x\]

Answer 30

right?

Answer 31

sstarica.. the original was 3x^2.
I agree with lokisan

Answer 32

oh , sorry my bad ^^"

Answer 33

is it clearer now ballards?

Answer 34

thank you guys that makes since, but what if you have a log problem like this...
f(x)=5x-xlnx

Answer 35

it is clear how to figure out the relative rate of change. Could someone help me figure out the f'(x) of that log then i maybe could get it from there

Answer 36

if f(x)=ln(x) f'(x)=1/x

Answer 37

same story, find the derivative then divide it by the original function ^_^ and you'll get :

f'(x) = 5 - [lnx + 1] ( use u'v + uv' to find the derivatice of xlnx)

then divide it by the original :
\[f'(x)/f(x) = (5-\ln|x|-1)/ 5x - xlnx\]

and try to simplify :) give it a try now.