SOLUTION: This is a calculus question: Please Please Help! Find all the points of inflection (concavity) f(x)= x^4-18x^2+4

Question 57283: This is a calculus question:
Please Please Help!
Find all the points of inflection (concavity)
f(x)= x^4-18x^2+4

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
f(x)= x^4-18x^2+4
f' = 4x^3-36x
f''=12x^2-36
Let f''=0:
12x^2-36=0
x^2-3=0
x=sqrt3 or x=-sqrt3
These are the candidate values:
Now, does f'' change sign at
x=sqrt3 or at x=-sqrt3
Recall f''=12x^2-36
As x approches sqrt3 from the left
f'' is negative because 12x^2 will be
less than 36.
As x approches sqrt3 from the right f''
is positive because 12x^2 will be greater
than 36.
So there is a point of inflection at x=sqrt3.
---------
The same logic will hold for x=-sqrt3 because
of the term 12x^2 .
You have two points of inflection: x=sqrt3 and
x=-sqrt3
Cheers,
Stan H.
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