Question

Hard Calculus Problem. Anyone up to the challenge?

Answer 1

How do I find the critical points? This is the second derivative.
\[\frac{ 288x(x^{-3}-4) }{ (x ^{3}+8)^{3} }\]

Answer 2

Not critical points but the concavity

Answer 3

@ganeshie8

Answer 4

you set the second derivative equal to 0 and solve x for the potential x values where the concavity changes, right?

Answer 5

yes but how would I set the equation equal to 0

Answer 6

like this : \[\frac{ 288x(x^{-3}-4) }{ (x ^{3}+8)^{3} }=0\]

Answer 7

yes but I do not know what to do after that

Answer 8

find where the numerator equals 0: \[288x\left(\frac{1}{x^3}-4\right)=0\]

Answer 9

what happened to the denominator

Answer 10

Solve for the numerator first, then work on the denominator, one step at a time.

Answer 11

Although, the denominator tells you where the function does not exist, or has vertical asymptotes.

Answer 12

would i divide 288x on both sides?

Answer 13

So solve them separately. \[288x=0\]\[\left(\frac{1}{x^3}-4\right) = 0\]

Answer 14

x=0, the other one you add 4 to both sides then do what?

Answer 15

original problem #14

Answer 16

http://www.math.rutgers.edu/~sferry/MA135F14/135-f14/Final2008.pdf

Answer 17

\[\frac{1}{x^3} = 4\]\[x^3 = \frac{1}{4}\]Take the cube root of both sides. and remember that \(4=2^2\)\[\large x=\sqrt[3]{\frac{1}{2^2}}\]\[x=\frac{1}{(2^2)^{1/3}} \implies x=\frac{1}{2^{2/3}}\implies 2^{-2/3}\]

Answer 18

how is x^3=1/4

Answer 19

I flipped 1/x^3, so I had to flip 4 to 1/4

Answer 20

but the answer you put doesnt match with the link that i posted

Answer 21

\(4^{1/3}\)? I am not sure how they got that.

Answer 22

it's fine ill ask one of my friends tomorrow

Answer 23

do you know where they got the -2 from?

Answer 24

oh from the denominator

Answer 25

increasing and decreasing points are found from finding the critical points of the first derivative.

Answer 26

you sure?

Answer 27

you set the derivative to 0? isnt that the local min/max

Answer 28

The vertical asymptote you are talking about is from solving for the denominator, you are right.

Answer 29

i am talking about when it increase/decrease

Answer 30

how do you find that?

Answer 31

you test values into your first derivative at the critical points (the little chart to the right)

Answer 32

which values

Answer 33

sorry for all these question i just have my finals tomorrow

Answer 34

I really appreciate the help

Answer 35

So do I...lol

Answer 36

for calc lol?

Answer 37

Calc 3.

Answer 38

freshmen?

Answer 39

no, sophomore.

Answer 40

so which values from the previous question

Answer 41

are you allowed a calculator?, because its pretty trivial with that. in any case, these notes are invaluable: http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx

Answer 42

Ok, so I'll answer this one, then I am off.

You find your vertical asymptote from the denominator once you find your first derivative. \[x^3 = -8\]\[x=-2\]

Answer 43

You then make the little chart like they have there to pick "test values, from the left and right side of the critical value, in this case -2, and test whether your function is increasing or decreasing by plugging values into the first derivative.

Answer 44

k