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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well lets see this dont make sence so i cant help sorry i wish i could

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What's the function you're looking at?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[ f(x) = x ^{4} + 6x ^{3}  24x ^{2} + 26\] I get the answer of x=1 and x=4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a calculus class?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh good. Didn't want to try to factor that ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did not know what she wanted because im in the 6th grade haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = 4x^3 + 18x^248x = 0\] I'm getting \(x \in \{1.8807, 0, 6.3807\} \) for the critical points.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this looks so confusing hehe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And of those, the only inflection point is the 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i think ill get it soon when i get in a different grade

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's ok moon, keep practicing and you'll get to calculus sooner than you think.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you take the 2nd derivative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i became a fan of you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f''(x) = 12x^2+36x  48 = 0\] \[\implies x^2 + 3x  4 = 0 \implies x \in \{1,4\}\] So actually 0 isn't an inflection point either.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since none of the critical points are also points at which f'' is 0 you don't have any points of inflection.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is what I get, too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that's true. Technically they are inflection points. I just typically think of inflection points as in the stationary kind. Those are nonstationary points of inflection because while the concavity does change, the tangent to the function is nonzero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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".

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I am kinds confussed with this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They 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.
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