At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

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

so when I plug I get - 20/3

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

correct

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