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

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Question 57281: Please Please Help!
Find all the points of inflection (concavity)
f(x)= x^4-18x^2+4
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
I am assuming that you are in Calculus. Points of inflection are the points on a graph that the function changes concavity (either from up to down or from down to up. You find the points of inflection by setting the second derivative equal to zero.
Find all the points of inflection.
f%28x%29=+x%5E4-18x%5E2%2B4 Take the first derivative.
f'(x)=4x^3-36x Take the derivative again, this is the second derivative.
f"(x)=12x^2-36 Set the derivative = 0, factor and solve.
0=12x%5E2-36
0=12%28x%5E2-3%29
0%2F12=x%5E2-3
0=x%5E2-3
3=x%5E2
%2B-sqrt%283%29=sqrt%28x%5E2%29
x=sqrt%283%29 and x=-sqrt%283%29
These are the x values for the points of inflection, you have to plug these values into the original f(x) to find the y values.
f%28sqrt%283%29%29=%28sqrt%283%29%29%5E4-18%28sqrt%283%29%29%5E2%2B4
f%28sqrt%283%29%29=9-18%283%29%2B4
f%28sqrt%283%29%29=9-54%2B4
f%28sqrt%283%29%29=-41 This point of inflection is (sqrt(3) ,-41).
:
f%28-sqrt%283%29%29=%28-sqrt%283%29%29%5E4-18%28-sqrt%283%29%29%5E2%2B4
f%28-sqrt%283%29%29=9-54%2B4
f%28-sqrt%283%29%29=-41 This point of inflection is (-sqrt(3),-41)
:
When you look at the graph you realize this is resonable. (I actually caught a mistake when I checked myself against the graph when I solved your problem.) Here's the graph:
graph%28300%2C200%2C-10%2C10%2C-80%2C6%2Cx%5E4-18x%5E2%2B4%29
Can you tell about when the graph looks like a cup is turning upside down, then right side up?
Happy Calculating!!!