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help_needed
 3 years ago
Find the linearization L(x) of the function at a.
f(x) = x^4 + 4x^2, a = −1
help_needed
 3 years ago
Find the linearization L(x) of the function at a. f(x) = x^4 + 4x^2, a = −1

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help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0I keep getting F(x) = f(x) = 12x  8 as my asnwer but it's wrong :(

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1is this right? \[L(x)=f(a)+(xa)f'(a) \]

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1so we have a=1 and\[L(x)=f(1)+(x+1)f'(1)\]how about this one ? make sense?

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1i think u got it from here :)

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0yeah but thats exactly what i do but the asnwer sumhow is wrong

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0f(1) = 5 and f'(1) = 12. So it becomes 5 + 12(x+1) = 12x + 8

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0and it keeps marking it wrong

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0idk what im doing wrong here

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0srry that shud be 12x 8

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0WOWW IM SUCHHH A FAILLURE...OK wow i cant beleive i got the calc right and algebra wrong

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0thank u lol i really cant beleive i got this wrong

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1no problem ..it happens sometime :)

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0ok but this one its not about the algebra. Im just cnfused how to do it. Do u mind helping if i post it

help_needed
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks im posting it as a dffnt question
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