Question

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?

Answer 1

Can you find the first derivative of f(x)

Answer 2

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

Answer 3

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

Answer 4

the answer says -24 for 3.5 tho :S

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

@ sabika13 did you get it? any question here?

Answer 9

use the quotient rule here

Answer 10

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

Answer 11

can follow from here?

Answer 12

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

Answer 13

@ sabika13 are you following now here?

Answer 14

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

Answer 15

what does d and v stand for :S

Answer 16

oh i see, are you in algebra class now?

Answer 17

advanced functions

Answer 18

yes, do you know limits?

Answer 19

limits of a function?

Answer 20

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

Answer 21

im not sure if idid.. or what that is

Answer 22

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

Answer 23

thanks

Answer 24

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

Answer 25

YW, good luck now