Question

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

Answer 1

start with
\[y=x^2(x-2)(x+2)\] so you see that you have 3 zeros, at x =-2,x=0,x=2

Answer 2

the zero at x = 0 has multiplicity 2, so the graph will touch the x - axis there but not cross it

Answer 3

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

Answer 4

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?

Answer 5

And this point can not be on the graph, this makes me confused

Answer 6

yes, for the second coordinate you evaluate the function at
\[\sqrt{\frac{2}{3}}\]

Answer 7

yes, so I should get 2 inflection points

Answer 8

it must be on the graph,
plug in the x, find the y

Answer 9

right
\[\pm\sqrt{\frac{2}{3}}\]

Answer 10

so when I plug I get - 20/3

Answer 11

for point of inflection we need only to equate first derivative of f(x) to 0.not the second

Answer 12

actually it is the second
first derivative gives relative max or min

Answer 13

it should be plugged in the original formula, right?

Answer 14

yes, original function
you have the x, you want the y

Answer 15

correct

Answer 16

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

Answer 17

but im sure that i can get the points of inflection from first derivative,not the second