## anonymous 5 years ago Locate all inflection points of:

1. anonymous

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

2. anonymous

What's the function you're looking at?

3. anonymous

$f(x) = x ^{4} + 6x ^{3} - 24x ^{2} + 26$ I get the answer of x=1 and x=-4

4. anonymous

This is a calculus class?

5. anonymous

yes

6. anonymous

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

7. anonymous

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

8. anonymous

$f'(x) = 4x^3 + 18x^2-48x = 0$ I'm getting $$x \in \{1.8807, 0, -6.3807\}$$ for the critical points.

9. anonymous

this looks so confusing hehe

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

did you take the 2nd derivative?

14. anonymous

ok thanks polpak

15. anonymous

i became a fan of you

16. anonymous

$f''(x) = 12x^2+36x - 48 = 0$ $\implies x^2 + 3x - 4 = 0 \implies x \in \{1,-4\}$ So actually 0 isn't an inflection point either.

17. anonymous

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

18. anonymous

Does that make sense?

19. anonymous

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

20. anonymous

That is what I get, too.

21. anonymous

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.

22. anonymous

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".

23. anonymous

So I am kinds confussed with this

24. anonymous

"kinda"

25. anonymous

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.

26. anonymous

i see now. thanks!!