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Mathematics 7 Online
OpenStudy (anonymous):

Can I get help with this Cal 1 prob

OpenStudy (anonymous):

where is it?

OpenStudy (anonymous):

OpenStudy (anonymous):

number 3

OpenStudy (anonymous):

This is unfeigned cheating...

OpenStudy (anonymous):

how so

OpenStudy (anonymous):

I am not blind, I can see that you are taking an online test.

OpenStudy (anonymous):

agree...

OpenStudy (anonymous):

this is a quiz not a test we take tests in person

OpenStudy (anonymous):

and we are allowed to get help on quizes as well as hw

OpenStudy (amistre64):

:) what have you tried so far? maybe i can help iron it out

OpenStudy (amistre64):

q3 looks greened out, which one of these are you actually needing help with?

OpenStudy (anonymous):

I got the points of inflection but it said it was wrong on the last one i did it asks similar questions that generally have close to the same answer

OpenStudy (anonymous):

3

OpenStudy (anonymous):

that my first choice just not sure if it is correct would like someone to verify

OpenStudy (amistre64):

i like working with product rule better that quotient rule; makes the mathing simpler to me

OpenStudy (anonymous):

^

OpenStudy (amistre64):

\[f(x)=x(9x^2-16)^{-1}\] \[f'(x)=x'(9x^2-16)^{-1}+x(9x^2-16)'^{-1}\]

OpenStudy (amistre64):

we need the 2nd derivative for concavity and infections

OpenStudy (amistre64):

\[f'(x)=x'(9x^2-16)^{-1}+x(9x^2-16)'^{-1}\] \[f''(x)=x''(9x^2-16)^{-1}+x'(9x^2-16)'^{-1}+x'(9x^2-16)'^{-1}+x(9x^2-16)''^{-1}\] \[f''(x)=x''(9x^2-16)^{-1}+2x'(9x^2-16)'^{-1}+x(9x^2-16)''^{-1}\]

OpenStudy (amistre64):

can you sort that out?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

OK i think i got it

OpenStudy (anonymous):

also i am completly confused on this one

OpenStudy (amistre64):

should come out to this in the end

OpenStudy (amistre64):

with that mess .... we find the zeros for top and bottom to test out

OpenStudy (anonymous):

Yea thats what i got on that one

OpenStudy (anonymous):

How do i do number 6 though completly confused.

OpenStudy (amistre64):

you get to the second derivative :)

OpenStudy (amistre64):

f(x) = cx^2 - 3x^-2 f'(x) = 2cx +6x^-3 f''(x) = 2c -18x^-4

OpenStudy (anonymous):

why did i have to post to this question lol notification crazy going on right now

OpenStudy (amistre64):

now when x = 4, this ideally should equal 0

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

f''(4) = 0 = 2c - 18/4^4 18/4^4 = 2c 9/4^4 = c

OpenStudy (amistre64):

4^4 = (2^2)^4 = 2^8 = 256 i believe

OpenStudy (anonymous):

then wat do i do after that

OpenStudy (amistre64):

you compare c = 9/256 with the answer options and see what matches ......

OpenStudy (anonymous):

ok cool i can do that

OpenStudy (amistre64):

lol

OpenStudy (anonymous):

also do u know what they are asking for this question

OpenStudy (amistre64):

im sure that i do

OpenStudy (amistre64):

do you recall what a derivative (f') tells us about a function?

OpenStudy (anonymous):

whether is conclaves down or not right

OpenStudy (amistre64):

close, thats f'' the 2nd derivative f' tells us the slope of the tangent line to any given point

OpenStudy (amistre64):

so the graph of a derivative is negative when the slope at a point is negative; 0 when we reach a min or max or sometimes an inflection, and postive when the slope is increasing

OpenStudy (anonymous):

oh ok see we havant really gone over how to get slopes from the graphs

OpenStudy (anonymous):

how can i do that

OpenStudy (amistre64):

im sure they give you options to choose from ....

OpenStudy (anonymous):

oh im sorry i dint let u see the options let me reload it.

OpenStudy (anonymous):

OpenStudy (amistre64):

|dw:1333414190749:dw| the slopes should map out something like this

OpenStudy (anonymous):

OpenStudy (amistre64):

|dw:1333414268730:dw|

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