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I think you have to derive here, or am I wrong? Since it's about rate of change, right?
no this is the problem
the derivative would be 3x^2 -6x
that's how you compute the relative rate of change, by deriving ^_^
im not understanding?
so do i take f'(x) and divide it by f(x)
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 :)
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
or am I mixing it up with something?
i got all that, but my teacher still counted 2 points off?
hmm, I'm missing a point aren't I?
sstarica, when you're done here, can you help dinaortega? I'm leaving soon.
I'll try my best loki :)
could you help me lokisan?
relative rate of change is f'(x)/f(x)
>_< thank you blexting
blexting so how do i get that from f(x)= x^3 -3x^2 being the original problem and f'(x)=3x^2 -6x
Sorry.. I don't know, but I knew the formula.. sstarica can you help?
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divide f'(x) with f(x)
so you'll get :\[f'(x)/f(x) = (3x^2-6x)/(x^3-2x^2)\]\[= 3x(x-2)/x^2(x-2)\] \[= 3/x\]
sstarica.. the original was 3x^2. I agree with lokisan
oh , sorry my bad ^^"
is it clearer now ballards?
thank you guys that makes since, but what if you have a log problem like this... f(x)=5x-xlnx
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
if f(x)=ln(x) f'(x)=1/x
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.