Question

When dealing with Deriviatives, and pulling information of an function's graph from said function analytically, if the 2nd derivative set equal to 0 is undefined/irrational, does that mean there is an asymptote along the graph?

Specifically, the problem I'm working on has a second derivative of "2 + 6x^(-4)". So, theoretically, x = (-3)^(1/4). This is undefined/ irrational, yet what does this mean? How do we know which values of x have the asymptote, IF that's what an undefined answer here means? :/ I need help with understanding on this!

Any and all help is greatly appreciated! :)

Answer 1

Also, does this mean that the graph of the function has no Inflection Points?

Answer 2

what is the original function?

Answer 3

y = ((x^4) + 1)/x^2

Answer 4

better known as \(f(x)=x^2+\frac{1}{x^2}\) ok

Answer 5

Yep, I get that. :)

Answer 6

so the second derivative is
\[\frac{6}{x^4}+2\]right?

Answer 7

Yes, that's what I got.

Answer 8

or as you had it
\[2+6x^{-4}\]

Answer 9

:)

Answer 10

the problem with using exponential notation with negative exponents is that you cannot really compute with them or solve with them

Answer 11

perhaps it was not clear how to solve
\[2+6x^{-4}=0\] but it is more clear with
\[2+\frac{6}{x^2}=0\]
or at least clearer

Answer 12

or rather \[2+\frac{6}{x^4}=0\]
it is not obvious that there is no solution

Answer 13

i meant to say "it is NOW obvious that there is no solution"

Answer 14

So...
2 + (6/x^4) = 0
-2
_________________
6/(x^4) = -2
(x^4)(6/(x^4)) = -2(x^4)
6 = -2x^4
__________
-2
x^4 = -3
Yeah, get how that is found! but what, conceptually, does it MEAN in terms of the graph ??

Answer 15

hold the phone

Answer 16

:) sorry it took me a while to type out that math.

Answer 17

Okay

Answer 18

you are doing tons too much work
whatever \(x\) is it is certain that \(x^4\) is positive, and since \(6\) is positive we know \(\frac{6}{x^4}\) is positive, so certainly \(2+\frac{6}{x^4}\) is positive and therefore it is not zero

Answer 19

so there is no solution
i noticed you wrote "irrational/undefined"
it is not "irrational" there is no real number solution
what this tells you is that the second derivative is always positive
what that says about the original function is that is always leans left, or as your math teacher probably says "concave up"

Answer 20

So... If the second derivative is not solvable for x (undefined) then the graph is always concave up, all the time?

Answer 21

no

Answer 22

if it is never zero, then it is always either positive or negative (assuming it is continuous)
that means the function is either always concave up or concave down, depending on whether the second derivative is positive or negative

Answer 23

*rgggg* ;)

Answer 24

Ahhhhh.... Okay. That makes beautiful, perfect sense.

Answer 25

clear right? never zero doesn't mean always positive, just in this case it was
it could also be always negative

Answer 26

:))) thank you so much!

Answer 27

think of a real simple example, like \(y=ax^2+bx+c\) where \(y''=2a\)

Answer 28

if \(a>0\) then it is always concave up, which you knew already since it is a parabola that opens up
but if \(a<0\) then it is always concave down

Answer 29

yw

Answer 30

a = 0 (?)

Answer 31

So, the I.P. Is at x = 0, and it's concave up?

Answer 32

then so is the second derivative, and you have a line

Answer 33

if \(a=0\) then \(y=ax^2+bx+c=bx+c\) a line which is neither concave up or down

Answer 34

Don't you plug the value of x from the 2nd deriviative into the original equation to find the y-value of the I.P.?

Answer 35

i don't know what an IP is

Answer 36

oh inflection point
nvm

Answer 37

Sorry

Answer 38

if the second derivative is never zero, then there is not inflection point

Answer 39

Okay

Answer 40

no inflection point is what i was trying to write
the inflection point is at the zero of the second derivative

Answer 41

Yep, I think I get it all now. Ha, I know I at least get it a heck of a lot better. :)

Answer 42

Is there an asymptote at x = 0 then?

Answer 43

@satellite73

Answer 44

for the original function?

Answer 45

Yeah in the graph of the original function.

Answer 46

yes, because that would make the denominator equal to zero

Answer 47

Okay.

Answer 48

So... are +/- 1 minima (1,2) & (-1.2) ?

Answer 49

i don't know, i didn't do it
hold on

Answer 50

yes

Answer 51

Alright.... And the , because it will approach 0, to infinity, it is increasing over the intervals: [-1, 0) U [1, 🔃)

Answer 52

increasing on \((-1,0)\) and \((1\infty)\) looks good
make the intervals open

Answer 53

Okay, sweet. Awesome, this is great, I've learned more in like, an hour online than I have in two weeks in Calc at school.... That's nice. :p thank you :)

Answer 54

your quite welcome
good luck

Answer 55

I have another problem, quite similar- would it be alright if I sort of continue into that one, and ask you questions again as they arise in my brain?

Answer 56

Haha... I am sorry, and feel free to say no, but what would you say the second deriviative of "y = x/((x^2) - 4)" is?

Answer 57

I used the quotient rule right out of the gate, is that what you'd do?

Answer 58

\[\frac{x}{x^2-4}\]?

Answer 59

Yep :)

Answer 60

\[\frac{2x^3+24x}{(x-4)^3}\] i think

Answer 61

typo there try
\[\frac{2x^3+24x}{(x^2-4)^3}\]