TheCalcHater
  • TheCalcHater
I need some calculus help...
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TheCalcHater
  • TheCalcHater
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Miss.Rose
  • Miss.Rose
I don't know calculus, however I know a website that can help you. http://www.mathway.com It's called Mathway. It really works!
TheCalcHater
  • TheCalcHater
@Miss.Rose I have tried it.... it wont work for this problem

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TheCalcHater
  • TheCalcHater
I am leaning towards B. Any opinions?
jim_thompson5910
  • jim_thompson5910
you agree that the volume of a cube formula is `V = s^3` right?
TheCalcHater
  • TheCalcHater
yes volume is s^3 or L*W*H
jim_thompson5910
  • jim_thompson5910
well L, W, H are all equal to s so that's how L*W*H = s*s*s = s^3
jim_thompson5910
  • jim_thompson5910
apply the derivative with respect to s on V = s^3 to get dV/ds = ???
TheCalcHater
  • TheCalcHater
right so how would i find the volume and surface area if there is room for error?
jim_thompson5910
  • jim_thompson5910
V = s^3 dV/ds = _______ (fill in the blank)
TheCalcHater
  • TheCalcHater
3
jim_thompson5910
  • jim_thompson5910
not just 3
TheCalcHater
  • TheCalcHater
3s?
jim_thompson5910
  • jim_thompson5910
more like 3s^2
jim_thompson5910
  • jim_thompson5910
hopefully you learned the power rule?
TheCalcHater
  • TheCalcHater
yeah i see now u wanted me to take the derivative
jim_thompson5910
  • jim_thompson5910
http://study.com/academy/lesson/power-rule-for-derivatives-examples-lesson-quiz.html
TheCalcHater
  • TheCalcHater
now that we have the first derivative do we need to find the second?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
we're going to use differentials
TheCalcHater
  • TheCalcHater
ok (and yeah i know the power rule :) )
jim_thompson5910
  • jim_thompson5910
dV/ds = 3s^2 dV = 3s^2*ds ... multiply both sides by `ds`
TheCalcHater
  • TheCalcHater
what is ds equal to?
jim_thompson5910
  • jim_thompson5910
s = 17 is the measured side length ds = 0.08 is the error in the measured side length
jim_thompson5910
  • jim_thompson5910
ds is the differential in s
TheCalcHater
  • TheCalcHater
oh ok
TheCalcHater
  • TheCalcHater
ok now what do i do?
jim_thompson5910
  • jim_thompson5910
dV is the differential in volume (aka the error in volume)
jim_thompson5910
  • jim_thompson5910
plug in s = 17 and ds = 0.08 and evaluate
TheCalcHater
  • TheCalcHater
1179.12
jim_thompson5910
  • jim_thompson5910
too large
jim_thompson5910
  • jim_thompson5910
dV = 3s^2*ds dV = 3*17^2*0.08 dV = _______________ (fill in the blank)
TheCalcHater
  • TheCalcHater
right when i plugged in and multiplied that is what i got
TheCalcHater
  • TheCalcHater
oh wait nvm
TheCalcHater
  • TheCalcHater
69.36
TheCalcHater
  • TheCalcHater
so it is d...
TheCalcHater
  • TheCalcHater
?
jim_thompson5910
  • jim_thompson5910
yes, let's also do the surface area as well A = surface area A = 6s^2 dA/ds = 12s dA = 12s*ds s = 17 and ds = 0.08 dA = 12s*ds dA = 12*17*0.08 dA = ???
TheCalcHater
  • TheCalcHater
16.32! Thank you so much. Do you think you can help me with a few more?
jim_thompson5910
  • jim_thompson5910
sure
TheCalcHater
  • TheCalcHater
ok there are three more i think i have a good start to cal 1
TheCalcHater
  • TheCalcHater
Also i appreciate you helping me understand instead of giving me the answers (i hate when people do that) :)
TheCalcHater
  • TheCalcHater
Any ideas?
jim_thompson5910
  • jim_thompson5910
one moment
TheCalcHater
  • TheCalcHater
ok
jim_thompson5910
  • jim_thompson5910
were you able to find f ' (x) ?
TheCalcHater
  • TheCalcHater
for which one?
jim_thompson5910
  • jim_thompson5910
the first one. It's better to post one at a time to avoid confusion like this
TheCalcHater
  • TheCalcHater
ok give me a second
TheCalcHater
  • TheCalcHater
x(2x+1)−1/2+(2x+1)1/2
TheCalcHater
  • TheCalcHater
for d. i kow there are not any horizontal asymptotes
jim_thompson5910
  • jim_thompson5910
`for d. i kow there are not any horizontal asymptotes` that is correct
TheCalcHater
  • TheCalcHater
and for a. I dont think there are any critical numbers when i put it into my calc it says error
jim_thompson5910
  • jim_thompson5910
I'm assuming you meant to write this? \[\Large f \ '(x) = x(2x+1)^{-1/2} + (2x+1)^{1/2}\]
TheCalcHater
  • TheCalcHater
yes
TheCalcHater
  • TheCalcHater
sorry the way i copied it from my calc didnt do it witth the ^
jim_thompson5910
  • jim_thompson5910
Let's factor out the GCF \[\Large f \ '(x) = x(2x+1)^{-1/2} + (2x+1)^{1/2}\] \[\Large f \ '(x) = (2x+1)^{1/2}\left(x(2x+1)^{-1} + 1\right)\]
TheCalcHater
  • TheCalcHater
ok
jim_thompson5910
  • jim_thompson5910
Then a bit of simplification \[\Large f \ '(x) = (2x+1)^{1/2}\left(x(2x+1)^{-1} + 1\right)\] \[\Large f \ '(x) = (2x+1)^{1/2}\left(\frac{x}{2x+1} + 1\right)\] \[\Large f \ '(x) = (2x+1)^{1/2}\left(\frac{x}{2x+1} + 1*\frac{2x+1}{2x+1}\right)\] \[\Large f \ '(x) = (2x+1)^{1/2}\left(\frac{x}{2x+1} + \frac{2x+1}{2x+1}\right)\] \[\Large f \ '(x) = (2x+1)^{1/2}\left(\frac{x+2x+1}{2x+1}\right)\] \[\Large f \ '(x) = (2x+1)^{1/2}\left(\frac{3x+1}{2x+1}\right)\] \[\Large f \ '(x) = \frac{(2x+1)^{1/2}}{2x+1}(3x+1)\] \[\Large f \ '(x) = \frac{(2x+1)^{1/2}}{(2x+1)^1}(3x+1)\] \[\Large f \ '(x) = (2x+1)^{1/2-1}(3x+1)\] \[\Large f \ '(x) = (2x+1)^{-1/2}(3x+1)\] \[\Large f \ '(x) = \frac{3x+1}{\sqrt{2x+1}}\] Hopefully you agree with all of these steps?
TheCalcHater
  • TheCalcHater
give me a second to check it
TheCalcHater
  • TheCalcHater
yep looks good
jim_thompson5910
  • jim_thompson5910
to find the critical values, you need to solve `f ' (x) = 0`
jim_thompson5910
  • jim_thompson5910
\[\Large f \ '(x) = \frac{3x+1}{\sqrt{2x+1}}\] \[\Large 0 = \frac{3x+1}{\sqrt{2x+1}}\] x = ???
TheCalcHater
  • TheCalcHater
x=-1/3
TheCalcHater
  • TheCalcHater
so a is -1/3 :)
jim_thompson5910
  • jim_thompson5910
ok and we also need to look to see where f ' (x) is undefined AND where f(x) is defined what happens when you set the denominator of f ' (x) equal to 0 and solve for x?
TheCalcHater
  • TheCalcHater
it is undefined
TheCalcHater
  • TheCalcHater
?
jim_thompson5910
  • jim_thompson5910
sqrt(2x+1) = 0 leads to x = ???
TheCalcHater
  • TheCalcHater
x=1/2
jim_thompson5910
  • jim_thompson5910
-1/2 actually
TheCalcHater
  • TheCalcHater
so the critical points are -1/3 and -1/2?
jim_thompson5910
  • jim_thompson5910
the domain of f is x >= -1/2 so this solution is in the domain of f
jim_thompson5910
  • jim_thompson5910
correct
TheCalcHater
  • TheCalcHater
ok how do we do part b?
jim_thompson5910
  • jim_thompson5910
draw out a number line and plot the following on it -1/2 = -0.5 -1/3 = -0.3333....
jim_thompson5910
  • jim_thompson5910
|dw:1449797807105:dw|
TheCalcHater
  • TheCalcHater
|dw:1449797814716:dw|
TheCalcHater
  • TheCalcHater
lol urs is sooo much better
jim_thompson5910
  • jim_thompson5910
we essentially have 2 regions to test region A: between -1/2 and -1/3 region B: between -1/3 and +infinity |dw:1449797873115:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1449797889588:dw|
TheCalcHater
  • TheCalcHater
to find?
jim_thompson5910
  • jim_thompson5910
what's a point between -1/2 and -1/3 on the number line?
TheCalcHater
  • TheCalcHater
in region a it can be anything
jim_thompson5910
  • jim_thompson5910
what's a point between -0.5 and -0.3333 on the number line? pick any decimal you want
TheCalcHater
  • TheCalcHater
-.4
jim_thompson5910
  • jim_thompson5910
sure that works |dw:1449798027685:dw|
jim_thompson5910
  • jim_thompson5910
plug x = -0.4 into `f ' (x)` and tell me what you get
TheCalcHater
  • TheCalcHater
1.64 approx
TheCalcHater
  • TheCalcHater
are you still here?
jim_thompson5910
  • jim_thompson5910
1.64 is incorrect
jim_thompson5910
  • jim_thompson5910
Try again \[\Large f \ '(x) = \frac{3x+1}{\sqrt{2x+1}}\] \[\Large f \ '(-0.4) = \frac{3(-0.4)+1}{\sqrt{2(-0.4)+1}}\] \[\Large f \ '(-0.4) = ???\]
TheCalcHater
  • TheCalcHater
oh sorry i put +.4
TheCalcHater
  • TheCalcHater
-.45
jim_thompson5910
  • jim_thompson5910
yep
TheCalcHater
  • TheCalcHater
on what next?
jim_thompson5910
  • jim_thompson5910
f ' is negative all throughout region A so the slopes of f are negative we have a decreasing interval |dw:1449798483764:dw|
jim_thompson5910
  • jim_thompson5910
now pick a value from region B plug it into f ' (x)
TheCalcHater
  • TheCalcHater
after this question i have to go unfortunatly do you think you can send the other two questions and their explainations to my email?
jim_thompson5910
  • jim_thompson5910
we can work on this when you get back
TheCalcHater
  • TheCalcHater
positive .4
jim_thompson5910
  • jim_thompson5910
0 is a better value since it's easiest to work with plug in x = 0 to get f ' (x) = ???
TheCalcHater
  • TheCalcHater
that will be -1.64
jim_thompson5910
  • jim_thompson5910
nope
TheCalcHater
  • TheCalcHater
1.64*
jim_thompson5910
  • jim_thompson5910
Try again \[\Large f \ '(x) = \frac{3x+1}{\sqrt{2x+1}}\] \[\Large f \ '(0) = \frac{3(0)+1}{\sqrt{2(0)+1}}\] \[\Large f \ '(0) = ???\]
TheCalcHater
  • TheCalcHater
I thought we were using .4??
TheCalcHater
  • TheCalcHater
oh nvm u suggesting using 0
TheCalcHater
  • TheCalcHater
1
jim_thompson5910
  • jim_thompson5910
oh I see now, yes f ' (0.4) = 1.64
jim_thompson5910
  • jim_thompson5910
it doesn't matter which x value you pick as long as it's in region B
TheCalcHater
  • TheCalcHater
lets stick with 0 :) it will be 1
TheCalcHater
  • TheCalcHater
what is this finding? the part b (the open intervals increasing or decreasing)?
jim_thompson5910
  • jim_thompson5910
|dw:1449798878369:dw|
TheCalcHater
  • TheCalcHater
ok ??
jim_thompson5910
  • jim_thompson5910
where are you stuck?
TheCalcHater
  • TheCalcHater
so at point 1/3 it is increasing?
jim_thompson5910
  • jim_thompson5910
what is the decreasing interval?
TheCalcHater
  • TheCalcHater
-1/2? so the critical points?
jim_thompson5910
  • jim_thompson5910
anywhere in region A, right?
TheCalcHater
  • TheCalcHater
yeah
jim_thompson5910
  • jim_thompson5910
why region A? because plugging ANY point from region A into the derivative function gives you a negative value negative derivative value ----> negative tangent line slope ----> f is decreasing
jim_thompson5910
  • jim_thompson5910
so the decreasing interval, in interval notation, is (-1/2, -1/3)
TheCalcHater
  • TheCalcHater
right and since it is a negative slope it is decreasing at any point
TheCalcHater
  • TheCalcHater
so the answer to part b would be anywhere in region a decreasing and -1/3 increasing?
TheCalcHater
  • TheCalcHater
because it says open intervals ??
jim_thompson5910
  • jim_thompson5910
your teacher won't know what "region A" means unless you provided the number line why not use interval notation like I just wrote?
TheCalcHater
  • TheCalcHater
oh i didnt see the interval sorry
TheCalcHater
  • TheCalcHater
and the increasing?
jim_thompson5910
  • jim_thompson5910
you tell me
TheCalcHater
  • TheCalcHater
(-1/3, infinity) ?
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
decreasing interval: (-1/2, -1/3) increasing interval (-1/3, +infinity)
jim_thompson5910
  • jim_thompson5910
so our f(x) function looks something like this |dw:1449799377525:dw|
TheCalcHater
  • TheCalcHater
yay I'm understanding this a bit better thank you.. now for part c. finding the relative extrema.
jim_thompson5910
  • jim_thompson5910
|dw:1449799393702:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1449799432150:dw|
TheCalcHater
  • TheCalcHater
gottcha
jim_thompson5910
  • jim_thompson5910
i'm sure at this point, you see where the relative extrema is
jim_thompson5910
  • jim_thompson5910
and what type of extrema it is
TheCalcHater
  • TheCalcHater
how do we find the relative extrema? is it just the max and min?
TheCalcHater
  • TheCalcHater
so -1/2? and concave?
jim_thompson5910
  • jim_thompson5910
relative extrema = either local max or local min
TheCalcHater
  • TheCalcHater
so then -1/2 and minimum
jim_thompson5910
  • jim_thompson5910
look at my last drawing
TheCalcHater
  • TheCalcHater
(-1/2,-1/3) ?
jim_thompson5910
  • jim_thompson5910
and refer back to the sign chart number line
jim_thompson5910
  • jim_thompson5910
you're just guessing at this point
jim_thompson5910
  • jim_thompson5910
what were the critical values again?
TheCalcHater
  • TheCalcHater
i just dont understand what you mean...
jim_thompson5910
  • jim_thompson5910
where is f ' (x) = 0 true? what x values?
TheCalcHater
  • TheCalcHater
-1/3 and -1/2
jim_thompson5910
  • jim_thompson5910
yes, and only -1/3 works because there's nothing to the left of -1/2
TheCalcHater
  • TheCalcHater
to answer the critical point question i am working on the f(x)=0 give me a second
jim_thompson5910
  • jim_thompson5910
notice how f ' changes sign as you pass through x = -1/3
TheCalcHater
  • TheCalcHater
right so that is a concavity point.... will that also be the min or max?
jim_thompson5910
  • jim_thompson5910
no, we don't know anything about concavity right now we haven't determined the second derivative
TheCalcHater
  • TheCalcHater
oh ok so that point is the turning point.... so what does that make it?
jim_thompson5910
  • jim_thompson5910
is it a local min? or local max?
TheCalcHater
  • TheCalcHater
it looks like the min because it is at the bottom portion and because the graphs max is infinitiy
jim_thompson5910
  • jim_thompson5910
if you didn't have the graph, how would you determine if it's a min or max?
TheCalcHater
  • TheCalcHater
plugging it into f(x) to see if it is the lowest or highest possibility
TheCalcHater
  • TheCalcHater
or id it f'(x)?
jim_thompson5910
  • jim_thompson5910
you'd use the first derivative test http://clas.sa.ucsb.edu/staff/lee/Max%20and%20Min's.htm right?
TheCalcHater
  • TheCalcHater
ok i see how they are doing it so are we going to use the shot gun method or do we have our answer?
jim_thompson5910
  • jim_thompson5910
`first derivative test` you have to scroll a bit don't pay attention to the shot gun method. I don't like that method. It's too vague and not accurate enough
TheCalcHater
  • TheCalcHater
so we are looking at the 2nd derivative test?
jim_thompson5910
  • jim_thompson5910
FIRST derivative test
TheCalcHater
  • TheCalcHater
ok so we have our first derivative.... then what (I'm still reading)
TheCalcHater
  • TheCalcHater
so we will plug in 0 to start?
jim_thompson5910
  • jim_thompson5910
I don't know what you mean. Honestly I'm not sure where you're stuck. Did you read through that portion of the article? where it starts with `first derivative test`
TheCalcHater
  • TheCalcHater
yes where it mentions the numbers to plug in i was asking is that what we are going to do.... are we going to start plugging in 0 for x or did we do that already?
jim_thompson5910
  • jim_thompson5910
you mean when they did f ' (0) = 9 ?
TheCalcHater
  • TheCalcHater
yep. what are we going to do? are we going to plug in x=0 for f'(x)?
TheCalcHater
  • TheCalcHater
but if i recall we did that already...
jim_thompson5910
  • jim_thompson5910
yeah we did. The article is just going over a different example
TheCalcHater
  • TheCalcHater
ok so how do we do ours?
TheCalcHater
  • TheCalcHater
i know ours should be a minimum.. i am not sure about what pount i thought it was -1/2
TheCalcHater
  • TheCalcHater
point*
jim_thompson5910
  • jim_thompson5910
why -1/2 ?
TheCalcHater
  • TheCalcHater
because that was the lowest point on the graph but u asked about not having the graph
jim_thompson5910
  • jim_thompson5910
how is it the lowest point when we've shown that region A is decreasing?
TheCalcHater
  • TheCalcHater
|dw:1449801027063:dw| that point here is it not -1/2?
TheCalcHater
  • TheCalcHater
can you set up the problem and then ill solve it?
jim_thompson5910
  • jim_thompson5910
I think I see the confusion (-1/2, -1/3) is NOT a point like you learn about in algebra 1 it is an interval start at -1/2, end at -1/3
jim_thompson5910
  • jim_thompson5910
it's unfortunate that the notation is the same for points and interval notation (open endpoints)
TheCalcHater
  • TheCalcHater
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh ok so it is like a line kinda?
TheCalcHater
  • TheCalcHater
ok so then how do we find the min?
jim_thompson5910
  • jim_thompson5910
|dw:1449801149139:dw|
TheCalcHater
  • TheCalcHater
so the minimum can be anywhere on there?
jim_thompson5910
  • jim_thompson5910
no, that region is decreasing it bottoms out when x = -1/3
jim_thompson5910
  • jim_thompson5910
the local min happens at x = -1/3 to find what the local min is, we plug x = -1/3 back into f(x) NOT f ' (x)
TheCalcHater
  • TheCalcHater
ok so where it bottoms out it should be the min? because it is the bottom of the U
jim_thompson5910
  • jim_thompson5910
yeah
TheCalcHater
  • TheCalcHater
ok thank you i am going to try the other 2 questions because i have to go.. i will be on tomorrow if i need more help :) how did u get so good at calc? do you teach it?
TheCalcHater
  • TheCalcHater
A. x=-1/3 and x=-1/2 B. decreasing interval is (-1/2, -1/3) the increasing interval is (-1/3, positive infinity) C. the relative extrema is -1/3 and is the minimum D. None.
jim_thompson5910
  • jim_thompson5910
when you plugged x = -1/3 into f(x), what did you get?

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