anonymous
  • anonymous
Locate all inflection points of:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
well lets see this dont make sence so i cant help sorry i wish i could
anonymous
  • anonymous
What's the function you're looking at?
anonymous
  • anonymous
\[ f(x) = x ^{4} + 6x ^{3} - 24x ^{2} + 26\] I get the answer of x=1 and x=-4

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anonymous
  • anonymous
This is a calculus class?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Oh good. Didn't want to try to factor that ;p
anonymous
  • anonymous
i did not know what she wanted because im in the 6th grade haha
anonymous
  • anonymous
\[f'(x) = 4x^3 + 18x^2-48x = 0\] I'm getting \(x \in \{1.8807, 0, -6.3807\} \) for the critical points.
anonymous
  • anonymous
this looks so confusing hehe
anonymous
  • anonymous
And of those, the only inflection point is the 0.
anonymous
  • anonymous
but i think ill get it soon when i get in a different grade
anonymous
  • anonymous
It's ok moon, keep practicing and you'll get to calculus sooner than you think.
anonymous
  • anonymous
did you take the 2nd derivative?
anonymous
  • anonymous
ok thanks polpak
anonymous
  • anonymous
i became a fan of you
anonymous
  • 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.
anonymous
  • anonymous
Since none of the critical points are also points at which f'' is 0 you don't have any points of inflection.
anonymous
  • anonymous
Does that make sense?
anonymous
  • 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
anonymous
  • anonymous
That is what I get, too.
anonymous
  • 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.
anonymous
  • 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".
anonymous
  • anonymous
So I am kinds confussed with this
anonymous
  • anonymous
"kinda"
anonymous
  • 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.
anonymous
  • anonymous
i see now. thanks!!

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