Question

I need some calculus help...

Answer 1

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!

Answer 2

@Miss.Rose I have tried it.... it wont work for this problem

Answer 3

I am leaning towards B. Any opinions?

Answer 4

you agree that the volume of a cube formula is `V = s^3` right?

Answer 5

yes volume is s^3 or L*W*H

Answer 6

well L, W, H are all equal to s

so that's how L*W*H = s*s*s = s^3

Answer 7

apply the derivative with respect to s on V = s^3 to get dV/ds = ???

Answer 8

right so how would i find the volume and surface area if there is room for error?

Answer 9

V = s^3

dV/ds = _______ (fill in the blank)

Answer 10

3

Answer 11

not just 3

Answer 12

3s?

Answer 13

more like 3s^2

Answer 14

hopefully you learned the power rule?

Answer 15

yeah i see now u wanted me to take the derivative

Answer 16

http://study.com/academy/lesson/power-rule-for-derivatives-examples-lesson-quiz.html

Answer 17

now that we have the first derivative do we need to find the second?

Answer 18

no

Answer 19

we're going to use differentials

Answer 20

ok (and yeah i know the power rule :) )

Answer 21

dV/ds = 3s^2

dV = 3s^2*ds ... multiply both sides by `ds`

Answer 22

what is ds equal to?

Answer 23

s = 17 is the measured side length
ds = 0.08 is the error in the measured side length

Answer 24

ds is the differential in s

Answer 25

oh ok

Answer 26

ok now what do i do?

Answer 27

dV is the differential in volume (aka the error in volume)

Answer 28

plug in s = 17 and ds = 0.08 and evaluate

Answer 29

1179.12

Answer 30

too large

Answer 31

dV = 3s^2*ds

dV = 3*17^2*0.08

dV = _______________ (fill in the blank)

Answer 32

right when i plugged in and multiplied that is what i got

Answer 33

oh wait nvm

Answer 34

69.36

Answer 35

so it is d...

Answer 36

?

Answer 37

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 = ???

Answer 38

16.32! Thank you so much. Do you think you can help me with a few more?

Answer 39

sure

Answer 40

ok there are three more i think i have a good start to cal 1

Answer 41

Also i appreciate you helping me understand instead of giving me the answers (i hate when people do that) :)

Answer 42

Any ideas?

Answer 43

one moment

Answer 44

ok

Answer 45

were you able to find f ' (x) ?

Answer 46

for which one?

Answer 47

the first one. It's better to post one at a time to avoid confusion like this

Answer 48

ok give me a second

Answer 49

x(2x+1)−1/2+(2x+1)1/2

Answer 50

for d. i kow there are not any horizontal asymptotes

Answer 51

`for d. i kow there are not any horizontal asymptotes`
that is correct

Answer 52

and for a. I dont think there are any critical numbers when i put it into my calc it says error

Answer 53

I'm assuming you meant to write this?

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

Answer 54

yes

Answer 55

sorry the way i copied it from my calc didnt do it witth the ^

Answer 56

Let's factor out the GCF
\[\Large f \ '(x) = x(2x+1)^{-1/2} + (2x+1)^{1/2}\]
\[\Large f \ '(x) = (2x+1)^{1/2}\left(x(2x+1)^{-1} + 1\right)\]

Answer 57

ok

Answer 58

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?

Answer 59

give me a second to check it

Answer 60

yep looks good

Answer 61

to find the critical values, you need to solve `f ' (x) = 0`

Answer 62

\[\Large f \ '(x) = \frac{3x+1}{\sqrt{2x+1}}\]
\[\Large 0 = \frac{3x+1}{\sqrt{2x+1}}\]
x = ???

Answer 63

x=-1/3

Answer 64

so a is -1/3 :)

Answer 65

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?

Answer 66

it is undefined

Answer 67

?

Answer 68

sqrt(2x+1) = 0 leads to x = ???

Answer 69

x=1/2

Answer 70

-1/2 actually

Answer 71

so the critical points are -1/3 and -1/2?

Answer 72

the domain of f is x >= -1/2
so this solution is in the domain of f

Answer 73

correct

Answer 74

ok how do we do part b?

Answer 75

draw out a number line and plot the following on it

-1/2 = -0.5
-1/3 = -0.3333....