Question

Find the possible points of inflection:
(x^3) +243/x

Answer 1

I have got the 2nd derivative but I don't think there is a point of inflection...

Answer 2

yeah

Answer 3

You don't need to graph it.

Answer 4

You know what concavity is, right?
(i.e. you know what concave up and concave down means, correct?)

Answer 5

yes

Answer 6

Very good.

Answer 7

In general, when is the function \(f(x)\) concave down?

Answer 8

when f''(x) is negative

Answer 9

Yes, good.

Answer 10

And when is this function \(f(x)\) concave up?

Answer 11

when f''(x) is positive

Answer 12

Yes.

Answer 13

So, (for any function \(f(x)\)), the points of inflection are the points where concavity of the function changes.

Answer 14

okk

Answer 15

i.e. \(\color{black}{\displaystyle x=c }\) is a point of inflection, IFF, \(\color{black}{\displaystyle f''(c)=0 }\) and if \(\color{black}{\displaystyle f''(x)}\) has different signs before \(x=c\) and after \(x=c\).

Answer 16

yeah but I am having trouble finding when f''(x) =0

Answer 17

Have you found \(f''(x)\)?

Answer 18

yes

Answer 19

Can you tell me what it is?

Answer 20

sure

Answer 21

\[6x +\frac{ 486 }{ x ^{3} }\]

Answer 22

Yes, very good.

Answer 23

\(\color{black}{\displaystyle f''(x)=6x+\frac{486}{x^3} }\)

Answer 24

Now, in order to find possible points of inflection, set \(\color{black}{\displaystyle f''(x)=0 }\).

Answer 25

\[x ^{4}=-81\]

Answer 26

thats as far as i got

Answer 27

but wouldn't you need "i"

Answer 28

(Why do I say possible, because if \(\color{black}{\displaystyle f''(c+\delta) }\) and \(\color{black}{\displaystyle f''(c-\delta) }\) are both negative or both positive, then, despite that \(\color{black}{\displaystyle f''(c)=0 }\), the point \(x=c\) is not an inflection, since the concavity of the function doesn't change.)

Answer 29

(For sufficiently small \(\delta\). ... In other words, we are (or would be) testing the concavity near x=c, to see if the concavity indeed changes.)

Answer 30

\(\color{black}{\displaystyle f''(x)=0 }\)

\(\color{black}{\displaystyle 6x+\frac{486}{x^3}=0 }\)

Answer 31

\(\color{black}{\displaystyle \frac{6x^4}{x^3}+\frac{486}{x^3}=0 }\)

\(\color{black}{\displaystyle \frac{6x^4+486}{x^3}=0 }\)

\(\color{black}{\displaystyle x^4=81 }\)

So, far you are correct in solving for x.

Answer 32

oh excuse me, -81.

Answer 33

Obviously, \(\color{black}{\displaystyle \nexists x\in \mathbb{R}: x^4<0}\)

Answer 34

i.e. \(\color{black}{\displaystyle x^4=-81}\) doesn't have any real-number solutions.

Answer 35

That means there isn't an \(x\) to satisfy \(f''(x)=0\).
(we aren't considering exclusively imaginary numbers)

Answer 36

um ok, so should I put on my paper that it isn't concave up or concave down at any interval?

Answer 37

Well, if there isn't a point \(x=c\) such that \(f''(c)=0\), then you definitely don't have any points of inflection.

Answer 38

can you help you justify why there is no points of inflection?

Answer 39

help me*

Answer 40

Yes, I can.

Answer 41

There are no point of inflection because (in this case) \(f''(x)\ne 0\) (for any \(x\)).

Answer 42

oh ok

Answer 43

any questions /

Answer 44

?

Answer 45

so how would I find concavity?

Answer 46

just graph the 2nd derivative?

Answer 47

Concave down iff \(\color{black}{\displaystyle f''(x)<0 }\).
Concave up iff \(\color{black}{\displaystyle f''(x)>0 }\).

Answer 48

but how can i find concavity if i can't find when f"(x)=0?

Answer 49

These two equations above, in fact do have a solution.

\(\large (\)Maybe there is a confusion as to how can my function \(f(x)\) have a solution to \(f''(x)>0\), and have a solution to \(f''(x)<0\), BUT not have a solution to \(x=0\)? (Doesn't this directly contradict the definition of an inflection point, which is defined to be the point where concavity changes, and we have concavity of different signs, but we don't have any points of inflection?!!

The answer to this is, that \(x=0\) would have been our point of inflection (and you would see this even better if you solved \(f''(x)>0\) and \(f''(x)<0\)), however, \(x=0\) is not part of the function's domain, because there is a vertical asymptote at \(x=0\).\(\large )\)

Answer 50

\(\color{black}{\displaystyle f''(x) =\frac{x^4+486}{x^3} }\)

and \(\color{black}{\displaystyle (x^4+486)>0 }\) for all \(x\), so whether \(f''(x)\) is negative or positive is going to depend entirely on \(x^3\).

Set, \(x^3>0\) and \(x^3<0\) to see where the function is concave up, and where the function is concave down.

Answer 51

An equivalent to this set up, which is as well the solution, set \(x>0\) to see where the function is concave up, and set \(x<0\) to see where the function is concave down.

(Why? Because \(x^3>0\) iff \(x>0\), and \(x^3<0\) iff \(x<0\).)

Answer 52

.... and there are the intervals where the function is concave down and up ...