## anonymous 5 years ago Find the relative rate of change fro each of the following functions. f(x)=x^3 -3x^2

1. anonymous

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

2. anonymous

no this is the problem

3. anonymous

the derivative would be 3x^2 -6x

4. anonymous

exactly

5. anonymous

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

6. anonymous

im not understanding?

7. anonymous

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

8. 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 :)

9. 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

10. anonymous

^_^ clearer?

11. anonymous

or am I mixing it up with something?

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

I'll try my best loki :)

16. anonymous

could you help me lokisan?

17. anonymous

Thanks.

18. anonymous

np

19. anonymous

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

20. anonymous

>_< thank you blexting

21. 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

22. anonymous

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

23. anonymous

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24. anonymous

divide f'(x) with f(x)

25. anonymous

and simplify

26. anonymous

x^3-6x?

27. anonymous

6x/x^3

28. anonymous

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

29. anonymous

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

30. anonymous

right?

31. anonymous

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

32. anonymous

oh , sorry my bad ^^"

33. anonymous

is it clearer now ballards?

34. anonymous

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

35. 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

36. anonymous

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

37. 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.