anonymous
  • anonymous
how do you find the local minimum and maximum points of f(x)=x^4-6x^2
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
myininaya
  • myininaya
1st step find first derivative 2nd step set first derivative=0 and solve for x 3rd step test to see if the function is changing from increasing to decreasing (=>local max) or decreasing to increasing (=>local min) If neither of those happen, then you do not have a local min/local max (the slopes have to change sign at the critical number)
anonymous
  • anonymous
what is the derivative?
anonymous
  • anonymous
Introduce a new variable, X = x^2. Now you can rewrite f(x) as X^2 - 6X, since X^2 = (x^2)^2 = x^4. Now you can easily find where the derivative of X^2 - 6X = 0. The solution is X=0, and X=6. Now change them back into x, using X=x^2 => sqrt(X) = x. So you have the critical points of f(x) are 0, and +/- sqrt(6). I think that's right... figure out which of those are minimum and maximum points by looking at second derivative.

Looking for something else?

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