chris215
  • chris215
-
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
familiar with derivatives ?
chris215
  • chris215
yeah
ganeshie8
  • ganeshie8
Good. Start by finding first and second derivatives of the given function

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

chris215
  • chris215
-4x^3+4x 2nd -12^2+4
ganeshie8
  • ganeshie8
\(f(x) = -x^4 + 2x^2\) \(f'(x)=-4x^3+4x\) \(f''(x)=-12^2+4\)
ganeshie8
  • ganeshie8
critical points occur when \(f'(x)=0\) : \[-4x^3+4x=0\] solve \(x\)
chris215
  • chris215
x= 0,1,-1
ganeshie8
  • ganeshie8
Yes, evaluate the second derivative at each of those critical values
chris215
  • chris215
ok i got 0
anonymous
  • anonymous
Plz Check second derivative it Should be -12x^2 + 4
anonymous
  • anonymous
Now just evaluate above second derivative at each point x= 0,1,-1 f''(x) = -12(x)^2 + 4
chris215
  • chris215
ok I got 0 again..

Looking for something else?

Not the answer you are looking for? Search for more explanations.