Help me with this Dervivative

[dumbasiankid]

ok i have to find the P of Inflection, the
eq:
(x^3+2x^2) / 2(x^2-4x-12)(x-3)
I found a hole at x=-2 so i got (x^2) / 2(x-3)(x-6)
when i took the der of that i got
F'(x)=(-9x^2-36x) / 2[(x-3)(x-6)]^2
The 2nd F''(x)=2(18x-36)-(9x^2-36x)2[(x-3)(x-6)] / [(x-3)(x-6)]^2

ok one of the hint is x^3-6x^2+36 is either F'(x) or F"(x) and find the zero of x^3-6x^2+36 so i am guessing this is the whole der or just the top of f' or f" but i cant get my answere to equal to this.

 

[arthur ohlsten]

I haven't solved the derivative, but the point of inflection is x= a negative number.
Let us plot the curve f[x]=x^2/{2[x-6][x-3}

a horizontal assymptote at y=1/2
a vertical assymptote at x=6 and x=3

starting at x=-oo y=1/2
as x increases y decreases and at x=0 y=0 double zero at x=0 so function rebounds
as x increases y increases and ix=+oo at x=3-
at x=3+ y=-oo
as x increases y increases having a local maximum and then y decreases as x increases ,and at x=6- y=-oo
at x=6+ y=+oo
as x increases y decreases and is assymptotic to y=1/2 as x-->oo

from the curve it is obvious the point of inflection is to the left of x=0

hope this helps
Arthur

 

[galactus]

Inflection points occur where f''(x)=0.

f(x)=\(\displaystyle \frac{(x^{3}+2x^{2})} { 2(x^{2}-4x-12)(x-3)}\)

f'(x)=\(\displaystyle \frac{-9x(x-4)}{2(x-6)^{2}(x-3)^{2}}\)

f''(x)=\(\displaystyle \frac{9(x^{3}-6x{2}+36)}{(x-6)^{3}(x-3)^{3}}=\frac{12}{(x-6)^{3}}-\frac{3}{(x-3)^{3}}\)

\(\displaystyle \frac{12}{(x-6)^{3}}=\frac{3}{(x-3)^{3}}\)


\(\displaystyle 4(x-3)^{3}=(x-6)^{3}\)


\(\displaystyle 3x^{3}-18x^{2}+108=0\)


\(\displaystyle x=-2.01724315176\)

 

[Unco]

Inflection points occur where f''(x)=0.

They do indeed. However, f''(x) = 0 must be backed up by a change in sign of f'(x) for there to be an inflexion point.

<-- General statement -->

 

[galactus]

Right you are, Unco. I should've elaborated.

 

[dumbasiankid]

The horizontal assymtope when i solve it out

x-> oo , y-> 1/2
x-> -oo , y-> -1/2

how did you get x approach x-> -oo , y-> 1/2??

 

[Gene]

Go back to the original equation and divide top & bottom by x^3 to get
(1+2/x)/(2*(1-4/x-12/x^2)(1-3/x)
When x->oo anything divided by it ->0 so you have
(1)/(2*1)=1/2