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.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

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

Did you use f'(2) and f'(-2) to come to your conclusion about the relative minimum?

Would I use f''(2)--plug is in to find the sign of the second derivative at that point?

I did use f'(-2) and f'(2)

if you use the second derivative you would use f"(-1) and f"(1)

use the critical points found from the first derivative to find out the concavity

yes Tbates is right

That gives me pos-->neg for each interval

sorry, neg-->pos for the first, pos--> pos for the second

The second one should be positive --> negative. At x = 2 the second derivative is -3/25

if you have a change in concavity.... means you have an inflection point

I am using first derivative ... is that wrong?

nope, the first derivative will do the trick for this problem.

I don't see where you got -3/25--my calculator says that f'(2) = .4

(1-2^2)/[(1+2^2)^2] is (1-4)/[(1+4)^2] = -3/(5^2)= -3/25

Exactly!

Thanks so much everyone!