suppose f(x)= x^4 +ax^2. What is the value of a if f has a local min at x=2?

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suppose f(x)= x^4 +ax^2. What is the value of a if f has a local min at x=2?

Mathematics
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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.

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suppose . [f(x)=x^{4}+ax ^{2.}\] What is the value of a if f has a local min at x=2?
im not sure if im supposed to find the derivative then set x = 2 then solve for a or something else..
If there is a local min at x=2, then f'(2)=0.

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If it has a local min, then we know the derivative at that point must be 0. So take the derivative, set x = 2, and set the derivative =0, then solve for a.
Which is essentially what Xavier said, but with more words ;p
okay, i did that.. the answer is -8 but i got -4
Did you get 4x^3+2ax=0 as the derivative?
yes
Substitute x=2 4(2)^2+4a=0
4(2)^3+4a=0 I mean
it would be 4(2)^3
yeah
4a=-32
ohhh. okay. wow.
thank soo much. im stupid..
Everyone has those moments :P
tag!!
tag

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