Question

32(3x^4-8x^2-16)= 0 Is 2nd derivative so this will give me my inflection points. help please

Answer 1

If that is your 2nd derivative, first solve the equation for all real numbers.

Answer 2

thats what I need help with

Answer 3

let u = x^2 to convetr this to a quadratic in u, then solve using
standard solution for roots of quadratic.

Answer 4

so I will have 3U^2-8U-16?

Answer 5

I get 4, -1.3333333.. does this mean x= sqrt4, -sqrt1.3333333?

Answer 6

That's the conversion. You have solved for U but U is x^2, so take the square root of the solutions for U to get x.

Answer 7

look at the graph and see if this makes sense.

Answer 8

doesnt look right on the graph at all

Answer 9

Quick tip: convert your original U to fraction.

Answer 10

huh?

Answer 11

3U^2-8U-16=0
Factor to get (U-4)(3U-4)=0
U=4 or U=4/3
So, x=2, sqrt(4/3), -2, -sqrt(4/3)

Answer 12

i should have just factored and ignored the quadratic on this one,right?

Answer 13

yes.

Answer 14

Do you know what to do from here on?

Answer 15

let me give you the entire 2nd derivative. all of that over (x^2-4)^2

Answer 16

actually, its 32(3x^4-3x^2-16)/(x^2-4)^2

Answer 17

Oh, then it gets easier.

Answer 18

-8x^2 not 3

Answer 19

Because the denominator can't be negative, x cannot by 2 nor -2

Answer 20

right I had that already

Answer 21

so it should just be what we found, but I checked and its not

Answer 22

So, the only possible x-values are sqrt(4/3) and -sqrt(4/3)
Now, the slope HAS to change its sign from the inflection point's left to right.

Answer 23

funny thing you say that. I should have known. the function is undefine at -2,2, right? So it does not change signs even thogh the concavity does.. so there are none...lol

Answer 24

in the original function.ughhh, these inflection points always trip me up!

Answer 25

Thanks for all of the help, people!

Answer 26

how about help determining its domain and range?

Answer 27

Domain=All real numbers except 2 and -2