S
  • S
How to sketch the graph of f(x)=x^2(x^2-4) including the inflection point?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
start with \[y=x^2(x-2)(x+2)\] so you see that you have 3 zeros, at x =-2,x=0,x=2
anonymous
  • anonymous
the zero at x = 0 has multiplicity 2, so the graph will touch the x - axis there but not cross it
anonymous
  • anonymous
to find the point of inflection you need the second derivative, \[f(x)=x^4-4x^2\] \[f'(x)=4x^3-8x\] \[f''(x)=12x^2-8\] set equal to zero and solve

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S
  • S
My main question is that, when i find the inflection point I put square root of 2/3 into the original formula and i get -20/3 right?
S
  • S
And this point can not be on the graph, this makes me confused
anonymous
  • anonymous
yes, for the second coordinate you evaluate the function at \[\sqrt{\frac{2}{3}}\]
S
  • S
yes, so I should get 2 inflection points
anonymous
  • anonymous
it must be on the graph, plug in the x, find the y
anonymous
  • anonymous
right \[\pm\sqrt{\frac{2}{3}}\]
S
  • S
so when I plug I get - 20/3
anonymous
  • anonymous
for point of inflection we need only to equate first derivative of f(x) to 0.not the second
anonymous
  • anonymous
actually it is the second first derivative gives relative max or min
S
  • S
it should be plugged in the original formula, right?
anonymous
  • anonymous
yes, original function you have the x, you want the y
S
  • S
correct
S
  • S
But in the end my inflection point comes up to be (- square root of 2/3 ; - 20/3) and (square root of 2/3 ; - 20/3) and my minimum points are (-2;-4) and (2;-4) and it seems that the point of inflection cant be on the graph
anonymous
  • anonymous
but im sure that i can get the points of inflection from first derivative,not the second

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