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

- anonymous

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

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- anonymous

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

- anonymous

no this is the problem

- anonymous

the derivative would be 3x^2 -6x

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## More answers

- anonymous

exactly

- anonymous

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

- anonymous

im not understanding?

- anonymous

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

- anonymous

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 :)

- anonymous

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

- anonymous

^_^ clearer?

- anonymous

or am I mixing it up with something?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I'll try my best loki :)

- anonymous

could you help me lokisan?

- anonymous

Thanks.

- anonymous

np

- anonymous

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

- anonymous

>_< thank you blexting

- anonymous

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

- anonymous

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

- anonymous

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- anonymous

divide f'(x) with f(x)

- anonymous

and simplify

- anonymous

x^3-6x?

- anonymous

6x/x^3

- anonymous

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

- anonymous

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

- anonymous

right?

- anonymous

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

- anonymous

oh , sorry my bad ^^"

- anonymous

is it clearer now ballards?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

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