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Mathematics 5 Online
OpenStudy (anonymous):

help another one:(

OpenStudy (anonymous):

OpenStudy (anonymous):

its wrong:(

OpenStudy (anonymous):

OpenStudy (vinicius1):

wait, is that a derivative?

OpenStudy (anonymous):

yes:(

OpenStudy (freckles):

First you need to find p' given what p is

OpenStudy (freckles):

p=f(x)g(x) what is p'?

OpenStudy (freckles):

hint: use product rule

OpenStudy (anonymous):

idk:(

OpenStudy (freckles):

You don't know product rule?

OpenStudy (freckles):

(uv)'=u'v+uv' is product rule u=u(x) and v=v(x)

OpenStudy (anonymous):

yes i do the formula to find derivatives right?

OpenStudy (anonymous):

but i don't know how to work with derivatives

OpenStudy (freckles):

So if p(x)=f(x)g(x) then p'(x)=f'(x)g(x)+f(x)g'(x) you are asked to find p'(4)

OpenStudy (freckles):

so this means you will need to know f'(4), g(4), f(4), and g'(4)

OpenStudy (freckles):

what is f(4) ?

OpenStudy (anonymous):

just plug in after that?

OpenStudy (anonymous):

3

OpenStudy (freckles):

f(4) is 3?

OpenStudy (freckles):

g(4) is 3

OpenStudy (freckles):

I think you looked at wrong function so what is f(4)

OpenStudy (vinicius1):

f(4) is 7

OpenStudy (freckles):

ok f'(4)=?

OpenStudy (freckles):

you just need to know the slope at x=4 of f

OpenStudy (freckles):

just find the slope of that line going from x=4 to x=6

OpenStudy (freckles):

use the slope formula \[f'(4)=\frac{f(6)-f(4)}{6-4}\]

OpenStudy (anonymous):

oh i see!!!!! but then it would equal to 0

OpenStudy (freckles):

no

OpenStudy (anonymous):

like she said above

OpenStudy (freckles):

f(6)=?

OpenStudy (freckles):

how are you guys getting f'(4) is 0?

OpenStudy (anonymous):

i have no clue! i think i am looking at this whole function wrong

OpenStudy (freckles):

to find f(6) look a long the x-axis for 6 then scroll up until you see f's graph what is the y that corresponds to x=6?

OpenStudy (anonymous):

yes that is what i was doing, you asked for f(4) which is 3 but then you said that was wrong?

OpenStudy (freckles):

f(4) is not 3

OpenStudy (anonymous):

now f(6) is 5

OpenStudy (anonymous):

1?

OpenStudy (anonymous):

for f(3)

OpenStudy (freckles):

f(6)=5 is correct but f(4) is between 6 and 8

OpenStudy (vinicius1):

f(4)=7

OpenStudy (freckles):

f is labeled as the black graph i think you looked at the blue graph

OpenStudy (anonymous):

i idid that at the beginning

OpenStudy (freckles):

\[f'(4)=\frac{f(6)-f(4)}{6-4}=\frac{5-7}{6-4}\]

OpenStudy (freckles):

yep and you did again too

OpenStudy (anonymous):

lol yes i saw

OpenStudy (anonymous):

-2

OpenStudy (anonymous):

OpenStudy (anonymous):

just gave me a new problem since i got 1st one wrong, sorry!

OpenStudy (anonymous):

we'll leave it for tomorrow if you like!

OpenStudy (freckles):

well you would need to find f(8) , g(8), f'(8) , g'(8)

OpenStudy (freckles):

You should be able to find f(8) and g(8) pretty easily the trick to find f'(8) and g'(8) is to think slope find the slope of the line surrounding (8,f(8)) and (8,g(8)) You need to use two points on those lines to find the slopes of those lines

OpenStudy (anonymous):

f(8) is 2?

OpenStudy (freckles):

you are getting the curves backwards again

OpenStudy (anonymous):

okay give me a second

OpenStudy (anonymous):

6

OpenStudy (anonymous):

?

OpenStudy (freckles):

yes f(8)=6

OpenStudy (freckles):

and g(8)=2

OpenStudy (freckles):

now you need to find f'(8) to do this just find the slope that (8,6) lies on

OpenStudy (freckles):

you need two points on that line

OpenStudy (freckles):

you already have one (8,6) can you tell me another point that is on that line (in which (8,6) occurs)?

OpenStudy (anonymous):

3,8?

OpenStudy (freckles):

(3,8) is on a different line part

OpenStudy (freckles):

|dw:1411104878176:dw| This little thing I drew is the line we are looking at we have one (8,6) but (3,8) is not on it

OpenStudy (freckles):

|dw:1411104909647:dw| you can use any two of those points to find the slope of that line

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