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S

  • 4 years ago

How to sketch the graph of f(x)=x^2(x^2-4) including the inflection point?

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  1. anonymous
    • 4 years ago
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    start with \[y=x^2(x-2)(x+2)\] so you see that you have 3 zeros, at x =-2,x=0,x=2

  2. anonymous
    • 4 years ago
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    the zero at x = 0 has multiplicity 2, so the graph will touch the x - axis there but not cross it

  3. anonymous
    • 4 years ago
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    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

  4. S
    • 4 years ago
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    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?

  5. S
    • 4 years ago
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    And this point can not be on the graph, this makes me confused

  6. anonymous
    • 4 years ago
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    yes, for the second coordinate you evaluate the function at \[\sqrt{\frac{2}{3}}\]

  7. S
    • 4 years ago
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    yes, so I should get 2 inflection points

  8. anonymous
    • 4 years ago
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    it must be on the graph, plug in the x, find the y

  9. anonymous
    • 4 years ago
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    right \[\pm\sqrt{\frac{2}{3}}\]

  10. S
    • 4 years ago
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    so when I plug I get - 20/3

  11. fazna
    • 4 years ago
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    for point of inflection we need only to equate first derivative of f(x) to 0.not the second

  12. anonymous
    • 4 years ago
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    actually it is the second first derivative gives relative max or min

  13. S
    • 4 years ago
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    it should be plugged in the original formula, right?

  14. anonymous
    • 4 years ago
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    yes, original function you have the x, you want the y

  15. S
    • 4 years ago
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    correct

  16. S
    • 4 years ago
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    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

  17. fazna
    • 4 years ago
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    but im sure that i can get the points of inflection from first derivative,not the second

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