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How to sketch the graph of f(x)=x^2(x^2-4) including the inflection point?
start with \[y=x^2(x-2)(x+2)\] so you see that you have 3 zeros, at x =-2,x=0,x=2
the zero at x = 0 has multiplicity 2, so the graph will touch the x - axis there but not cross it
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
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?
And this point can not be on the graph, this makes me confused
yes, for the second coordinate you evaluate the function at \[\sqrt{\frac{2}{3}}\]
yes, so I should get 2 inflection points
it must be on the graph, plug in the x, find the y
right \[\pm\sqrt{\frac{2}{3}}\]
for point of inflection we need only to equate first derivative of f(x) to 0.not the second
actually it is the second first derivative gives relative max or min
it should be plugged in the original formula, right?
yes, original function you have the x, you want the y
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
but im sure that i can get the points of inflection from first derivative,not the second