Question

completing the square to find inflection points?

Answer 1

I have 12x^2-24x+8 as my second derivative and im completing the square to find the inflection points. I minused 8 from each side and got 144 as the number to complete the square the only problem is it isnt a perfect one.its 12(x-2)(x+12)

Answer 2

how do i get rid of 136. since its now 12(x-2)(x+12)=136

Answer 3

What's your original function? @skay

Answer 4

I'm too lazy to integrate it -.-

Answer 5

\[F(x)=x ^{4}-4x ^{3}+4x ^{2}\]

Answer 6

its exponents 4, 3, 2. sorry thats tiny.

Answer 7

I attempted to use the quadratic formula but its not giving reasonable numbers for my graph. i think im missing a step.

Answer 8

I know you can use the Cubic formula but either of these methods is way easier

Answer 9

Ok so inflection points occur where the concavity changes, or f''(x) changes sign.
To check if the concavity changes, we need to check where f''(x) = 0. If it does, then that x-value is a point of inflection.\[f''(x)=12x^2-24x+8\]\[12x^2-24x+8=0 \rightarrow 4(3x^2-6x+2)=0 \rightarrow 3x^2-6x+2=0 \implies x = 0.423,1.577\]And those are your inflection points. I hope you understand. If you have any questions, just ask.

@skay

Answer 10

what... how did you get the points?

Answer 11

I checked where f''(x) = 0. The x-values that you get for f''(x) = 0 are the inflection points.

Answer 12

Are you a bit confused with that?

Answer 13

Right... so you subtracted 2 from each side and then?

Answer 14

What do you mean I subtracted? You don't have to do anything. You just find the second derivative, which is a quadratic, and find its roots through quadratic formula. The roots are your inflection points.

Answer 15

Oh gotcha. you use the roots in the formula.

Answer 16

all i needed! thanks.

Answer 17

Sorry I should call it a conjecture before it's approved of.

Answer 18

I thought setting f"(x)=0 to find inflection points was pretty standard?
Sorry most of this stuff is still over my head.

Answer 19

Dw, you'll get used to it.