anonymous
  • anonymous
Can I get help with this Cal 1 prob
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
where is it?
anonymous
  • anonymous
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anonymous
  • anonymous
number 3

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anonymous
  • anonymous
This is unfeigned cheating...
anonymous
  • anonymous
how so
anonymous
  • anonymous
I am not blind, I can see that you are taking an online test.
anonymous
  • anonymous
agree...
anonymous
  • anonymous
this is a quiz not a test we take tests in person
anonymous
  • anonymous
and we are allowed to get help on quizes as well as hw
amistre64
  • amistre64
:) what have you tried so far? maybe i can help iron it out
amistre64
  • amistre64
q3 looks greened out, which one of these are you actually needing help with?
anonymous
  • 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
anonymous
  • anonymous
3
anonymous
  • anonymous
that my first choice just not sure if it is correct would like someone to verify
amistre64
  • amistre64
i like working with product rule better that quotient rule; makes the mathing simpler to me
anonymous
  • anonymous
^
amistre64
  • amistre64
\[f(x)=x(9x^2-16)^{-1}\] \[f'(x)=x'(9x^2-16)^{-1}+x(9x^2-16)'^{-1}\]
amistre64
  • amistre64
we need the 2nd derivative for concavity and infections
amistre64
  • amistre64
\[f'(x)=x'(9x^2-16)^{-1}+x(9x^2-16)'^{-1}\] \[f''(x)=x''(9x^2-16)^{-1}+x'(9x^2-16)'^{-1}+x'(9x^2-16)'^{-1}+x(9x^2-16)''^{-1}\] \[f''(x)=x''(9x^2-16)^{-1}+2x'(9x^2-16)'^{-1}+x(9x^2-16)''^{-1}\]
amistre64
  • amistre64
can you sort that out?
anonymous
  • anonymous
yes
anonymous
  • anonymous
OK i think i got it
anonymous
  • anonymous
also i am completly confused on this one
amistre64
  • amistre64
should come out to this in the end
1 Attachment
amistre64
  • amistre64
with that mess .... we find the zeros for top and bottom to test out
anonymous
  • anonymous
Yea thats what i got on that one
1 Attachment
anonymous
  • anonymous
How do i do number 6 though completly confused.
amistre64
  • amistre64
you get to the second derivative :)
amistre64
  • amistre64
f(x) = cx^2 - 3x^-2 f'(x) = 2cx +6x^-3 f''(x) = 2c -18x^-4
anonymous
  • anonymous
why did i have to post to this question lol notification crazy going on right now
amistre64
  • amistre64
now when x = 4, this ideally should equal 0
anonymous
  • anonymous
ok
amistre64
  • amistre64
f''(4) = 0 = 2c - 18/4^4 18/4^4 = 2c 9/4^4 = c
amistre64
  • amistre64
4^4 = (2^2)^4 = 2^8 = 256 i believe
anonymous
  • anonymous
then wat do i do after that
amistre64
  • amistre64
you compare c = 9/256 with the answer options and see what matches ......
anonymous
  • anonymous
ok cool i can do that
amistre64
  • amistre64
lol
anonymous
  • anonymous
also do u know what they are asking for this question
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amistre64
  • amistre64
im sure that i do
amistre64
  • amistre64
do you recall what a derivative (f') tells us about a function?
anonymous
  • anonymous
whether is conclaves down or not right
amistre64
  • amistre64
close, thats f'' the 2nd derivative f' tells us the slope of the tangent line to any given point
amistre64
  • 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
anonymous
  • anonymous
oh ok see we havant really gone over how to get slopes from the graphs
anonymous
  • anonymous
how can i do that
amistre64
  • amistre64
im sure they give you options to choose from ....
anonymous
  • anonymous
oh im sorry i dint let u see the options let me reload it.
anonymous
  • anonymous
1 Attachment
amistre64
  • amistre64
|dw:1333414190749:dw| the slopes should map out something like this
anonymous
  • anonymous
1 Attachment
amistre64
  • amistre64
|dw:1333414268730:dw|
anonymous
  • anonymous
that looks like A
amistre64
  • amistre64
umm, no further down; looks rather like an M
amistre64
  • amistre64
|dw:1333414363164:dw|
anonymous
  • anonymous
M theres no m lol
amistre64
  • amistre64
the graph "LOOKS LIKE" an M
anonymous
  • anonymous
Maybe c then
amistre64
  • amistre64
"M"aybe ;)
1 Attachment
anonymous
  • anonymous
one more question do you know to to find the conclave with the intervals
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