Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

  • S

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

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

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

  • 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
And this point can not be on the graph, this makes me confused
yes, for the second coordinate you evaluate the function at \[\sqrt{\frac{2}{3}}\]
  • S
yes, so I should get 2 inflection points
it must be on the graph, plug in the x, find the y
right \[\pm\sqrt{\frac{2}{3}}\]
  • S
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
  • S
it should be plugged in the original formula, right?
yes, original function you have the x, you want the y
  • S
correct
  • 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
but im sure that i can get the points of inflection from first derivative,not the second

Not the answer you are looking for?

Search for more explanations.

Ask your own question