Question

Locate all inflection points of:

Answer 1

well lets see this dont make sence so i cant help sorry i wish i could

Answer 2

What's the function you're looking at?

Answer 3

\[ f(x) = x ^{4} + 6x ^{3} - 24x ^{2} + 26\]

I get the answer of x=1 and x=-4

Answer 4

This is a calculus class?

Answer 5

yes

Answer 6

Oh good. Didn't want to try to factor that ;p

Answer 7

i did not know what she wanted because im in the 6th grade haha

Answer 8

\[f'(x) = 4x^3 + 18x^2-48x = 0\]

I'm getting \(x \in \{1.8807, 0, -6.3807\} \) for the critical points.

Answer 9

this looks so confusing hehe

Answer 10

And of those, the only inflection point is the 0.

Answer 11

but i think ill get it soon when i get in a different grade

Answer 12

It's ok moon, keep practicing and you'll get to calculus sooner than you think.

Answer 13

did you take the 2nd derivative?

Answer 14

ok thanks polpak

Answer 15

i became a fan of you

Answer 16

\[f''(x) = 12x^2+36x - 48 = 0\]
\[\implies x^2 + 3x - 4 = 0 \implies x \in \{1,-4\}\]

So actually 0 isn't an inflection point either.

Answer 17

Since none of the critical points are also points at which f'' is 0 you don't have any points of inflection.

Answer 18

Does that make sense?

Answer 19

from what i am reading in my book the inflection points is a point where the concavity changes. when i test some points i come up with\[x <-4 = positive \]\ \[x > 1 = pos\]\[-4 < x < 1 = negative \]
so it looks like it changes from positive at -4 to negative then postive again at 1

Answer 20

That is what I get, too.

Answer 21

Yes that's true. Technically they are inflection points. I just typically think of inflection points as in the stationary kind. Those are non-stationary points of inflection because while the concavity does change, the tangent to the function is non-zero.

Answer 22

ok for a question on is problem I need to"show inflections point(s) and explain in complete sentence why these number are inflection points and not critical points".

Answer 23

So I am kinds confussed with this

Answer 24

"kinda"

Answer 25

They are inflection points because the curvature changes (goes from positive to negative and then negative to positive). They are not critical points because the slope of the tangent to the curve at those points is not 0.

Answer 26

i see now. thanks!!