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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

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exactly

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)

^_^ clearer?

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?

Thanks.

np

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?

divide f'(x) with f(x)

and simplify

x^3-6x?

6x/x^3

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

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

right?

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

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

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