Question

Hi, can you help me find the inflection point(s) of this function
f(x)=(x^3)/((x^2)-4)

I found -2, 0 and 2 as my answer but it's wrong. Please help

Thank you :)

Answer 1

inflection points are where f''(x) =0.

Answer 2

I don't see any way to simplify the f(x) for the first derivative, so I would just go ahead and use a quotient rule. Or, re-write the x^2-4 thas on the bottom, as (x^2-4)^(-1) and then differentiate it with product rule.

Answer 3

what was your first derivative?

Answer 4

Actually I found my second derivative with online calculator

Answer 5

...

Answer 6

why calculator, why not by hand?

Answer 7

Because it's too difficult for me to find the second derivative.

Answer 8

It is not that difficult.

Answer 9

you can use the product rule for second derivative.

Answer 10

I mean re-writing the bottom as (x^2-4)^(-2)

Answer 11

The point is that you know HOW to find the derivative, so do you know how to find the first one without a calculator?

Answer 12

I think I do, but I always get a wrong answer. So really no I don't

Answer 13

Maybe because you are getting the rules incorrectly:)
\(\LARGE\color{black}{ \frac{d}{dx}\left(\begin{matrix} \frac{F(x)}{G(x)} \\ \end{matrix}\right) =\frac{g(x)~\color{red}{f \prime (x)}~-\color{red}{g \prime (x)}~f (x)}{ \left[ ~g(x)~ \right] ^2 } }\)

Answer 14

lets do it with your function, now,

Answer 15

tell me first 2 derivatives:
1) x^2-4
2) x^3

okay?

Answer 16

1) 2x = g'(x)
2) 3x^2 = f'(x)

Answer 17

good.

Answer 18

I got disconnected again

Answer 19

\(\LARGE\color{black}{ \frac{d}{dx}\left(\begin{matrix} \frac{\color{blue}{f(x)}}{\color{red}{g(x)}} \\ \end{matrix}\right) =\frac{\color{red}{g(x)}~\color{blue}{f \prime (x)}~-\color{red}{g \prime (x)}~\color{blue}{f (x)}}{ \left[ ~\color{red}{g(x)}~ \right] ^2 } }\)

\(\LARGE\color{black}{ \frac{d}{dx}\left(\begin{matrix} \frac{\color{blue}{x^3}}{\color{red}{x^2-4}} \\ \end{matrix}\right) =\frac{\color{red}{(x^2-4)}~\color{blue}{3x^2}~-\color{red}{2x}~\color{blue}{(x^3)}}{ \left[ ~\color{red}{(x^2-4)}~ \right] ^2 } }\)

Answer 20

Can you simplify this?

Answer 21

no need simplifying the bottom, just do the top part.

Answer 22

you can just draw it, (as long as I can read it).

Answer 23

or type the simplified result for the top that you get, here.

Answer 24

\[(3x^4-12x^2-2x^4)/(x^2-4)^2\]

Answer 25

yes, and what cancels on the top? subtract like terms, and you get?

Answer 26

x^2

Answer 27

where is this coming from?

Answer 28

jjst x^2 on the top??

Answer 29

*just

Answer 30

No wait

Answer 31

k

Answer 32

x^4

Answer 33

Because 3x^4-2x^4

Answer 34

x^4-12x^2

Answer 35

yes, exactly, so,

\(\LARGE\color{black}{ \frac{d}{dx}\left(\begin{matrix} \frac{\color{blue}{x^3}}{\color{red}{x^2-4}} \\ \end{matrix}\right) =\frac{x^4-12x^2 }{ \left[ ~\color{red}{(x^2-4)}~ \right] ^2 } }\)

Answer 36

now we need to find the second derivative.

Answer 37

tell me first (again same thing) the derivative off:
1) x^4 -12x^2
2) (x^2-4)^2 .

Answer 38

should we be using the quotient rule and chain rule?

Answer 39

we will have to use chain rule in either case, but we have a choice betwen a quotient rule, or product rule with negative exponenets.

Answer 40

1) 4x^3-24x
2) 2x*2(x^2-4)

Answer 41

yes. that is right.

Answer 42

We will use a product rule okay?

Answer 43

ok

Answer 44

\(\LARGE\color{black}{ \frac{d}{dx} \left[\color{white}{\frac{a}{b}}(x^4-12x^2)~(x^2-4)^{-2}~ \right] }\) this is what we will need to find.