The function y=x^4+bx^2+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

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The function y=x^4+bx^2+8x+1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

Mathematics
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so f ' (x) = 0 and f ' ' (x) = 0 for the same x . so f '(x) = f '' (x) f ' = 4x^3 + 2bx + 8 = f '' (x) = 12x^2 + 2b
but that still doesn't give me the value of b... do i set it equal to 0?
you have to solve that equation

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we have two equations 4x^3+2bx+8=0 and 12x^2+2b=0 I solve the bottom one for b and replace the b in the first equation with -6x^2
4x^3 + 2bx + 8= 12x^2 + 2b
oh thats smarter myininya has it
my way involves a bit of work
if you do this correctly you should get x=1 so now we can find b since b=-6x^2 then b=-6(1)^2=-6
any questions?
got it thank you!! :)
:)
i'm going to keep posting questions so feel free to answer :) lol... i have my ap test on wednesday so i'm reviewing a lot
good luck
thanks
and you could have done it cantorset's way but it is a bit more thinking down his path
i will show you if you want. i will do it on paper and scan it and post it
I'll try working on it in a bit, but I like the way you taught me better, seems a lot easier.
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wow.. looks like a lot of work :)
lol

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