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sabika13 3 years ago for the functions f(x)= 2x/(x-3), compare the slopes of the tangents i) at the points where x=3.5 and x=20 how do i do this?

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1. imron07

Can you find the first derivative of f(x)

2. sabika13

for x=3.5, it is 14 and for x=20 is 2.36

3. imron07

That's the slope of tangents at that points. (If you did find f'(x) correctly :))

4. sabika13

the answer says -24 for 3.5 tho :S

5. imron07

Tho? :? $f'(x)=\frac{2(x-3)-2x}{(x-3)^2}\\f'(3.5)=-24$

6. sabika13

though*.. I dont get it.. what did you do?

7. imron07

I thought it was some new math symbol. If f(x)=u/v then f'(x)=(u'v-uv')/v^2 Here, u=2x u'=2 v=x-3 v'=1

8. mark_o.

@ sabika13 did you get it? any question here?

9. mark_o.

use the quotient rule here

10. mark_o.

d(u/v)dx=(v du/dx - u dv/dx)/v^2

11. mark_o.

can follow from here?

12. mark_o.

for y= 2x/(x-3), let u =2x and v=(x-3) now do the quotient rule

13. mark_o.

@ sabika13 are you following now here?

14. sabika13

ivenever heard of the quotient rule.. or imforgetting..

15. sabika13

what does d and v stand for :S

16. mark_o.

oh i see, are you in algebra class now?

17. sabika13

advanced functions

18. mark_o.

yes, do you know limits?

19. mark_o.

limits of a function?

20. mark_o.

ok advance functions, im sure you already done limits of a function

21. sabika13

im not sure if idid.. or what that is

22. mark_o.

$\lim _{x->0}\frac{ f(x+ \Delta x )-f(x) }{ \Delta x }$

23. sabika13

thanks

24. mark_o.

do that limit as x approach to zero and you will arrived at f'(x)=-6/(x-3)^2 then you plug in x=3.5 and x=20 that is f'(3.5)=-24 and f'(20)=-0.021 and for f(3.5)=14 and f(20)= 3.36

25. mark_o.

YW, good luck now

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