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 3 years ago
solve this to get find global extreme values:
f(x)=x^48x^2+4x+2 at [(20/25),(64/25)]
 3 years ago
solve this to get find global extreme values: f(x)=x^48x^2+4x+2 at [(20/25),(64/25)]

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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0Why do you specify a point?

suju101
 3 years ago
Best ResponseYou've already chosen the best response.0yep!! extrema values fall in the range so it is specified.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0can't solve the cubic is all?

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0find criticle points check criticle points and boundary points

suju101
 3 years ago
Best ResponseYou've already chosen the best response.0i know that but i couldn't solve the equation

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0D[x^48x^2+4x+2 ]>= 4x^3 16x+4=0

imranmeah91
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=4x%5E3%2016x%2B4%3D0&t=crmtb01

suju101
 3 years ago
Best ResponseYou've already chosen the best response.0and how to solve that eqn to get value of x

sam_unleashed
 3 years ago
Best ResponseYou've already chosen the best response.0x^34x+1=0...then.. main problem is , to solve for x...
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